Skip to main content
Back to ELECTRIC POTENTIAL AND CAPACITANCE

Your Progress

0 / 45 Answered
0
Correct
0
Incorrect
0%
Accuracy
NEET • PHYSICS • ELECTRIC POTENTIAL AND CAPACITANCE

Electric potential and capacitance practice · Easy

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

The practice player uses single-correct-answer MCQs. Chapter quizzes help you review a topic; they are not full-length NEET mock papers.

Take a chapter quiz →
Showing 1-30 of 45 questions

Preparing practice and saved progress…

1 Easy
1 of 45Question permalink
Application +4 | -1
A capacitor has air as dielectric medium and two conducting plates of area 12 cm212 \mathrm{~cm}^2 and they are 0.6 cm0.6 \mathrm{~cm} apart. When a slab of dielectric having area 12 cm212 \mathrm{~cm}^2 and 0.6 cm0.6 \mathrm{~cm} thickness is inserted between the plates, one of the conducting plates has to be moved by 0.2 cm0.2 \mathrm{~cm} to keep the capacitance same as in previous case. The dielectric constant of the slab is: (Given ε0=8.834×10−12 F/m)\left(\text{Given }\varepsilon_0=8.834 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right)
View answer and explanation
Correct option: A
1.5
Option A (1.5) is correct because the dielectric constant can be found by equating the initial capacitance to the final capacitance after inserting the dielectric and moving the plate.
Solution approach
  1. Correct Answer + Core Reason: 1.5. The dielectric constant is found by equating the initial and final capacitances.

    Step-by-Step Solution:

    Step 1: Identify the given information: Area A=12cm2=12×10−4m2A = 12 \mathrm{cm}^2 = 12 \times 10^{-4} \mathrm{m}^2, initial distance d=0.6cm=0.6×10−2md = 0.6 \mathrm{cm} = 0.6 \times 10^{-2} \mathrm{m}, and the plate is moved by 0.2cm=0.2×10−2m0.2 \mathrm{cm} = 0.2 \times 10^{-2} \mathrm{m}.

    Step 2: The initial capacitance is given by C=ε0AdC = \frac{\varepsilon_0 A}{d}.

    Step 3: After inserting the dielectric and moving the plate, the new distance is d′=d+0.2×10−2m=0.8×10−2md' = d + 0.2 \times 10^{-2} \mathrm{m} = 0.8 \times 10^{-2} \mathrm{m}. The capacitance with the dielectric is C′=Kε0AdC' = \frac{K \varepsilon_0 A}{d}. The remaining air gap is d−0.6×10−2=0d - 0.6 \times 10^{-2} = 0. The effective distance is d′−t=0.8×10−2−0.6×10−2=0.2×10−2d'-t = 0.8 \times 10^{-2} - 0.6 \times 10^{-2} = 0.2 \times 10^{-2}. The new capacitance is C′=ε0Ad′−t+t/K=ε0A0.8×10−2−0.6×10−2+(0.6×10−2)/KC' = \frac{\varepsilon_0 A}{d' - t + t/K} = \frac{\varepsilon_0 A}{0.8 \times 10^{-2} - 0.6 \times 10^{-2} + (0.6 \times 10^{-2})/K}.\nSince the capacitance remains the same, C=C′C = C', so ε0A0.6×10−2=ε0A0.2×10−2+(0.6×10−2)/K\frac{\varepsilon_0 A}{0.6 \times 10^{-2}} = \frac{\varepsilon_0 A}{0.2 \times 10^{-2} + (0.6 \times 10^{-2})/K}.

    Step 4: Simplify the equation: 0.6×10−2=0.2×10−2+(0.6×10−2)/K0.6 \times 10^{-2} = 0.2 \times 10^{-2} + (0.6 \times 10^{-2})/K.
    0.4×10−2=(0.6×10−2)/K0.4 \times 10^{-2} = (0.6 \times 10^{-2})/K.
    K=0.6×10−20.4×10−2=1.5K = \frac{0.6 \times 10^{-2}}{0.4 \times 10^{-2}} = 1.5.

    Step 5: Final Answer: K=1.5K = 1.5.

    Why other options are wrong:
    - Option B: 1.33 is incorrect because it doesn't satisfy the capacitance equality condition.
    - Option C: 0.66 is incorrect because it's less than 1 and not physically possible in this scenario.
    - Option D: 1 is incorrect because it implies the dielectric has no effect, which contradicts the problem statement.

    Key Concepts:
    - Capacitance of a parallel plate capacitor: C=ε0AdC = \frac{\varepsilon_0 A}{d}
    - Effect of dielectric on capacitance: C′=KCC' = K C

    Memory Aids:
    - Remember that inserting a dielectric increases capacitance.

    Quick Tips:
    - Pay attention to the changes in distance when a dielectric is inserted and the plates are moved.
2 Easy
2 of 45Question permalink
Analyzing +4 | -1
Assertion (A) : From the relation C=qVC=\frac{q}{V}. We can say that, if more charge qq is given to a conductor, its capacitance should increase.

Reason (R) : Ratio qV\frac{q}{V} will remain constant for a given conductor.
View answer and explanation
Correct option: D
If Assertion is false but the Reason is true.
Option D (If Assertion is false but the Reason is true) is correct because capacitance is a property of the conductor and does not change with charge, but the ratio of charge to voltage remains constant.
Solution approach
  1. Correct Answer + Core Reason: If Assertion is false but the Reason is true. Capacitance is a property of the conductor and does not change with charge.

    Step-by-Step Solution:

    Step 1: Analyze the Assertion: The assertion states that if more charge is given to a conductor, its capacitance should increase. This is incorrect.

    Step 2: Analyze the Reason: The reason states that the ratio qV\frac{q}{V} will remain constant for a given conductor. This is correct. Capacitance is defined as C=qVC = \frac{q}{V}, and for a given conductor, C is constant.

    Step 3: Determine the relationship: The assertion is false, and the reason is true.

    Step 4: Final Answer: If Assertion is false but the Reason is true.

    Why other options are wrong:
    - Option A (If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.): This is incorrect because the assertion is false.
    - Option B (If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.): This is incorrect because the assertion is false.
    - Option C (If Assertion is true, but the Reason is false.): This is incorrect because the reason is true.

    Key Concepts:
    - Capacitance
    - Charge and potential relationship

    Memory Aids:
    - C=qVC = \frac{q}{V} (Capacitance = Charge / Voltage)

    Quick Tips:
    - Capacitance is a property of the conductor and does not depend on the charge or voltage.
3 Easy
3 of 45Question permalink
Application +4 | -1
A parallel plate condenser with oil between the plates (dielectric constant of oil K=2K = 2) has a capacitance C.C. If the oil is removed, then capacitance of the capacitor becomes
View answer and explanation
Correct option: D
C2\frac{C}{2}
Option D (C2\frac{C}{2}) is correct because removing the oil (dielectric) reduces the capacitance by a factor equal to the dielectric constant.
Solution approach
  1. Correct Answer + Core Reason: C2\frac{C}{2}. Removing the dielectric reduces capacitance by a factor of K.

    Step-by-Step Solution:

    Step 1: Recall the effect of a dielectric on capacitance.

    Step 2: The capacitance with a dielectric is C=KC0C = KC_0, where C0C_0 is the capacitance without the dielectric and KK is the dielectric constant.

    Step 3: Given K=2K = 2, we have C=2C0C = 2C_0. Therefore, C0=C2C_0 = \frac{C}{2}.

    Step 4: When the oil is removed, the capacitance becomes C0=C2C_0 = \frac{C}{2}.

    Why other options are wrong:
    - Option A: 2C\sqrt 2 C is incorrect because the capacitance decreases when the oil is removed.
    - Option B: 2C2C is incorrect because this is the capacitance *with* the oil.
    - Option C: C2\frac{C}{{\sqrt 2 }} is incorrect because the capacitance decreases by a factor of 2, not 2\sqrt{2}.

    Key Concepts:
    - Effect of dielectric on capacitance

    Memory Aids:
    - Remember that a dielectric increases capacitance.

    Quick Tips:
    - When a dielectric is removed, divide the capacitance by the dielectric constant.
4 Easy
4 of 45Question permalink
Application +4 | -1
The electric potential VV at any point (x,y,z)(x, y, z) in space is given according to relation V=2x2VV=2 x^2 {V}. The electric field intensity at the point (2m,0,2m)(2 \mathrm{m}, 0,2 \mathrm{m}) will be −E-E (volt/metre) then the value of EE is
View answer and explanation
Correct option: A
8
The correct answer is 8 because the electric field is the negative gradient of the potential, and evaluating this at the given point yields E = 8 V/m.
Solution approach
  1. Correct Answer + Core Reason: 8. The electric field is the negative gradient of the potential.

    Step-by-Step Solution:

    Step 1: Recall the relationship between electric field and electric potential: E⃗=−∇V=−(∂V∂xi^+∂V∂yj^+∂V∂zk^)\vec{E} = -\nabla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right).

    Step 2: Given V=2x2V = 2x^2, calculate the partial derivatives: ∂V∂x=4x\frac{\partial V}{\partial x} = 4x, ∂V∂y=0\frac{\partial V}{\partial y} = 0, ∂V∂z=0\frac{\partial V}{\partial z} = 0.

    Step 3: Therefore, E⃗=−4xi^\vec{E} = -4x \hat{i}.

    Step 4: Evaluate the electric field at the point (2 m,0,2 m)(2 \text{ m}, 0, 2 \text{ m}): E⃗=−4(2)i^=−8i^ V/m\vec{E} = -4(2) \hat{i} = -8 \hat{i} \text{ V/m}.

    Step 5: The magnitude of the electric field is E=∣−8∣=8 V/mE = |-8| = 8 \text{ V/m}.

    Why other options are wrong:
    - Option B: 0 is incorrect because the electric field is non-zero at the given point.
    - Option C: 0 is incorrect because the electric field is non-zero at the given point.
    - Option D: 0 is incorrect because the electric field is non-zero at the given point.

    Key Concepts:
    - Electric field and electric potential relationship
    - Gradient of a scalar field

    Memory Aids:
    - Remember that the electric field points in the direction of decreasing potential.

    Quick Tips:
    - Always remember to take the negative gradient of the potential to find the electric field.
5 Easy
5 of 45Question permalink
Application +4 | -1
A capacitor of capacitance 9 nF9 \mathrm{~nF} having dielectric slab of εr=2.4,{\varepsilon _r}=2.4, dielectric strength 20 MV/m20 \mathrm{~MV} / \mathrm{m} and P.D.=20 VP.D. = 20 \mathrm{~V} calculate area of plates.
View answer and explanation
Correct option: B
4.2×10−4 m24.2 \times 10^{-4} \mathrm{~m}^2
Option B is correct. The area of the plates is calculated using the formula C = (ε₀εᵣA)/d and V = Ed.
Solution approach
  1. Correct Answer + Core Reason: 4.2×10−4 m24.2 \times 10^{-4} \mathrm{~m}^2. Calculate area using capacitance, dielectric constant, and potential difference.

    Step-by-Step Solution:

    Step 1: Given: C=9 nF=9×10−9 FC = 9 \mathrm{~nF} = 9 \times 10^{-9} \mathrm{~F}, εr=2.4{\varepsilon _r}=2.4, E=20 MV/m=20×106 V/mE = 20 \mathrm{~MV} / \mathrm{m} = 20 \times 10^6 \mathrm{~V/m}, V=20 VV = 20 \mathrm{~V}.

    Step 2: We know that C=ε0εrAdC = \frac{{\varepsilon _0 \varepsilon _r A}}{d} and V=EdV = Ed, so d=VE=2020×106=10−6 md = \frac{V}{E} = \frac{20}{20 \times 10^6} = 10^{-6} \mathrm{~m}.

    Step 3: Substitute the value of d into the capacitance equation: 9×10−9=(8.854×10−12)(2.4)A10−69 \times 10^{-9} = \frac{(8.854 \times 10^{-12})(2.4)A}{10^{-6}}.

    Step 4: Solve for A: A=9×10−9×10−6(8.854×10−12)(2.4)=9×10−152.12496×10−11=4.235×10−4 m2≈4.2×10−4 m2A = \frac{9 \times 10^{-9} \times 10^{-6}}{(8.854 \times 10^{-12})(2.4)} = \frac{9 \times 10^{-15}}{2.12496 \times 10^{-11}} = 4.235 \times 10^{-4} \mathrm{~m}^2 \approx 4.2 \times 10^{-4} \mathrm{~m}^2.

    Why other options are wrong:
    - Option A: This is incorrect. It might be obtained by using an incorrect formula or making a calculation error.
    - Option C: This is incorrect. It might be obtained by using an incorrect formula or making a calculation error.
    - Option D: This is incorrect. It might be obtained by using an incorrect formula or making a calculation error.

    Key Concepts:
    - Capacitance of a parallel plate capacitor
    - Dielectric constant
    - Electric field and potential difference

    Memory Aids:
    - C=ε0εrAdC = \frac{{\varepsilon _0 \varepsilon _r A}}{d}, V=EdV = Ed.

    Quick Tips:
    - Remember to use consistent units.
6 Easy
6 of 45Question permalink
Analyzing +4 | -1
Assertion (A) : Two capacitors are connected in series with a battery. Energy stored across them is in inverse ratio of their capacity.

Reason (R): U=12qVU=\frac{1}{2} q V or U∝qVU \propto q V.
View answer and explanation
Correct option: B
If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
Option B (If both Assertion and Reason are true but Reason is not the correct explanation of Assertion) is correct. The energy stored is inversely proportional to capacitance because the charge is the same for capacitors in series, and U=q2/2CU = q^2/2C.
Solution approach
  1. Correct Answer + Core Reason: If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. In series, charge is constant, so energy is inversely proportional to capacitance, U=q2/2CU = q^2/2C.

    Step-by-Step Solution:

    Step 1: Analyze the Assertion.\nIn a series connection, the charge qq is the same for all capacitors. The energy stored in a capacitor is given by U=12q2CU = \frac{1}{2} \frac{q^2}{C}. Therefore, U∝1CU \propto \frac{1}{C}, meaning the energy stored is inversely proportional to the capacitance. The assertion is true.

    Step 2: Analyze the Reason.\nThe reason states U=12qVU = \frac{1}{2}qV or U∝qVU \propto qV. This is a correct formula for the energy stored in a capacitor. However, in a series connection, the voltage VV is not the same for all capacitors. The charge qq is the same. Therefore, this reason doesn't explain why the energy is inversely proportional to the capacitance in a series connection.

    Step 3: Determine if the Reason is the correct explanation for the Assertion.\nThe reason is true, but it does not correctly explain the assertion. The correct explanation is that in a series connection, the charge is constant, and U=12q2CU = \frac{1}{2} \frac{q^2}{C}, so U∝1CU \propto \frac{1}{C}.

    Why other options are wrong:
    - Option A: Incorrect. The reason is not the correct explanation for the assertion.
    - Option C: Incorrect. The reason is true.
    - Option D: Incorrect. The assertion is true.

    Key Concepts:
    - Capacitors in series
    - Energy stored in a capacitor

    Memory Aids:
    - In series, charge is constant.
    - Energy formulas: U=12CV2=12qV=12q2CU = \frac{1}{2}CV^2 = \frac{1}{2}qV = \frac{1}{2} \frac{q^2}{C}

    Quick Tips:
    - Remember that in series connections, charge is constant, while in parallel connections, voltage is constant.
7 Easy
7 of 45Question permalink
Remembering +4 | -1
A powerful machine used for producing high positive potentials is
View answer and explanation
Correct option: D
Van de graaff generator.
Option D is correct because a Van de Graaff generator is specifically designed to produce high positive potentials.
Solution approach
  1. Correct Answer + Core Reason: Van de graaff generator. Van de Graaff generators are designed to create high electrostatic potentials.

    Step-by-Step Solution:

    Step 1: Understand the function of each device.

    Step 2: A lightning conductor protects buildings from lightning strikes by providing a path to ground.

    Step 3: An electric motor converts electrical energy into mechanical energy.

    Step 4: A galvanometer measures small electric currents.

    Step 5: A Van de Graaff generator uses a moving belt to accumulate electric charge on a hollow metal globe, creating very high electrostatic potentials.

    Why other options are wrong:
    - Option A: A lightning conductor protects from high voltage but doesn't produce it.
    - Option B: An electric motor uses electricity, it doesn't produce high potentials.
    - Option C: A galvanometer measures current, it doesn't produce high potentials.

    Key Concepts:
    - Van de Graaff generator
    - Electrostatic potential

    Memory Aids:
    - Van de Graaff: High Voltage

    Quick Tips:
    - Remember the primary function of each device.
8 Easy
8 of 45Question permalink
Application +4 | -1
Two identical capacitors have same capacitance CC. One of them is charged to the potential VV and other to the potential 2V2 V. The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system is:
View answer and explanation
Correct option: A
14CV2\frac{1}{4} {CV}^2
Option A (14CV2\frac{1}{4} {CV}^2) is correct because it represents the decrease in energy after the capacitors are connected, calculated by finding the difference between the initial and final energies.
Solution approach
  1. Correct Answer + Core Reason: 14CV2\frac{1}{4} {CV}^2. The energy loss is the difference between initial and final energies.

    Step-by-Step Solution:

    Step 1: Calculate the initial energy of the first capacitor: U1=12CV2U_1 = \frac{1}{2}CV^2.

    Step 2: Calculate the initial energy of the second capacitor: U2=12C(2V)2=12C(4V2)=2CV2U_2 = \frac{1}{2}C(2V)^2 = \frac{1}{2}C(4V^2) = 2CV^2.

    Step 3: Calculate the total initial energy: Ui=U1+U2=12CV2+2CV2=52CV2U_i = U_1 + U_2 = \frac{1}{2}CV^2 + 2CV^2 = \frac{5}{2}CV^2.

    Step 4: Calculate the total charge before connection: Q1=CVQ_1 = CV and Q2=C(2V)=2CVQ_2 = C(2V) = 2CV. Total charge Q=Q1+Q2=CV+2CV=3CVQ = Q_1 + Q_2 = CV + 2CV = 3CV.

    Step 5: Calculate the equivalent capacitance after connection: Ceq=C+C=2CC_{eq} = C + C = 2C.

    Step 6: Calculate the final potential difference: Vf=QCeq=3CV2C=32VV_f = \frac{Q}{{{C_{eq}}}} = \frac{{3CV}}{{2C}} = \frac{3}{2}V.

    Step 7: Calculate the final energy: Uf=12CeqVf2=12(2C)(32V)2=C(94V2)=94CV2U_f = \frac{1}{2}C_{eq}V_f^2 = \frac{1}{2}(2C)(\frac{3}{2}V)^2 = C(\frac{9}{4}V^2) = \frac{9}{4}CV^2.

    Step 8: Calculate the decrease in energy: ΔU=Ui−Uf=52CV2−94CV2=104CV2−94CV2=14CV2\Delta U = U_i - U_f = \frac{5}{2}CV^2 - \frac{9}{4}CV^2 = \frac{{10}}{4}CV^2 - \frac{9}{4}CV^2 = \frac{1}{4}CV^2.

    Why other options are wrong:
    - **Option B (34CV2\frac{3}{4} {CV}^2): This is incorrect because it doesn't account for the correct final potential and energy after the capacitors are connected.
    -     Option C (2CV22 {CV}^2): This is incorrect because it doesn't account for the correct final potential and energy after the capacitors are connected.
    -     Option D (12CV2\frac{1}{2} {CV}^2): This is incorrect because it doesn't account for the correct final potential and energy after the capacitors are connected.

    Key Concepts:**
    - Energy stored in a capacitor: U=12CV2U = \frac{1}{2}CV^2
    - Charge on a capacitor: Q=CVQ = CV
    - Capacitance in parallel: Ceq=C1+C2C_{eq} = C_1 + C_2

    Memory Aids:
    - Remember to calculate the total charge and equivalent capacitance after connection.

    Quick Tips:
    - Always calculate the initial and final energies separately before finding the difference.
9 Easy
9 of 45Question permalink
Remembering +4 | -1
If E1E_1 be the electric field strength of a short dipole at a point on its axial line and E2E_2 that on the equatorial line at the same distance, then
View answer and explanation
Correct option: D
E1=2E2E _1=2 E _2
Option D (E1=2E2E _1=2 E _2) is correct because the electric field strength on the axial line of a short dipole is twice that on the equatorial line at the same distance.
Solution approach
  1. Correct Answer + Core Reason: E1=2E2E _1=2 E _2. The electric field on the axial line is twice the electric field on the equatorial line for a short dipole.

    Step-by-Step Solution:

    Step 1: Electric field on the axial line of a short dipole: E1=14πϵ02pr3E_1 = \frac{{1}}{{4\pi \epsilon_0}} \frac{{2p}}{{r^3}}, where pp is the dipole moment and rr is the distance from the center of the dipole.

    Step 2: Electric field on the equatorial line of a short dipole: E2=14πϵ0pr3E_2 = \frac{{1}}{{4\pi \epsilon_0}} \frac{{p}}{{r^3}}

    Step 3: Comparing E1E_1 and E2E_2: E1=2E2E_1 = 2E_2

    Why other options are wrong:
    - Option A: E1≠E2E _1 \neq E _2: They are not equal.
    - Option B: E1=E2E_1=E_2: The axial field is twice the equatorial field.
    - Option C: E2=2E1E _2=2 E _1: Incorrect relationship. The axial field is twice the equatorial field.

    Key Concepts:
    - Electric field due to a dipole
    - Axial and equatorial lines of a dipole

    Memory Aids:
    - Axial field is stronger than equatorial field.

    Quick Tips:
    - Remember the formulas for electric fields on the axial and equatorial lines of a dipole.
10 Easy
10 of 45Question permalink
Application +4 | -1
The difference in the effective capacitance of two equal capacitors when joined in parallel and series is 3  μF.3\,\,\mu F. The value of each capacitor is
View answer and explanation
Correct option: B
2  μF2\,\,\mu F
Option B (2  μF2 \,\,\mu F) is correct because it satisfies the condition that the difference in effective capacitance in parallel and series is 3  μF3 \,\,\mu F.
Solution approach
  1. Correct Answer + Core Reason: 2  μF2 \,\,\mu F. The difference between parallel and series capacitance is 3 μF\mu F.

    Step-by-Step Solution:

    Step 1: Define the capacitance of each capacitor as CC.

    Step 2: Calculate the equivalent capacitance in parallel: Cp=C+C=2CC_p = C + C = 2C.

    Step 3: Calculate the equivalent capacitance in series: Cs=C×CC+C=C22C=C2C_s = \frac{{C \times C}}{{C + C}} = \frac{{C^2}}{{2C}} = \frac{C}{2}.

    Step 4: Set up the equation for the difference in capacitance: Cp−Cs=3  μFC_p - C_s = 3 \,\,\mu F, which means 2C−C2=3  μF2C - \frac{C}{2} = 3 \,\,\mu F.

    Step 5: Solve for CC: 4C−C2=3  μF\frac{{4C - C}}{2} = 3 \,\,\mu F, so 3C2=3  μF\frac{{3C}}{2} = 3 \,\,\mu F, and C=2  μFC = 2 \,\,\mu F.

    Why other options are wrong:
    - **Option A (1  μF1 \,\,\mu F):** If C=1  μFC = 1 \,\,\mu F, then Cp=2  μFC_p = 2 \,\,\mu F and Cs=0.5  μFC_s = 0.5 \,\,\mu F. The difference is 1.5  μF1.5 \,\,\mu F, not 3  μF3 \,\,\mu F.
    - **Option C (3  μF3 \,\,\mu F):** If C=3  μFC = 3 \,\,\mu F, then Cp=6  μFC_p = 6 \,\,\mu F and Cs=1.5  μFC_s = 1.5 \,\,\mu F. The difference is 4.5  μF4.5 \,\,\mu F, not 3  μF3 \,\,\mu F.
    - **Option D (4  μF4 \,\,\mu F):** If C=4  μFC = 4 \,\,\mu F, then Cp=8  μFC_p = 8 \,\,\mu F and Cs=2  μFC_s = 2 \,\,\mu F. The difference is 6  μF6 \,\,\mu F, not 3  μF3 \,\,\mu F.

    Key Concepts:
    - Capacitance in parallel: Cp=C1+C2C_p = C_1 + C_2
    - Capacitance in series: 1Cs=1C1+1C2\frac{1}{{{C_s}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}

    Memory Aids:
    - Parallel: Capacitances add directly.
    - Series: Reciprocals add.

    Quick Tips:
    - Always calculate the equivalent capacitance for both parallel and series combinations before finding the difference.
11 Easy
11 of 45Question permalink
Application +4 | -1
An electric dipole is made up of two equal and opposite charges of 2×10−6C2 \times 10^{-6} {C} at a distance of 3 cm3 {~cm}. If this is kept in an electric field of 2×1052 \times 10^5 N/C{N} / {C}, then the maximum torque acting on the dipole is
View answer and explanation
Correct option: A
0.012
Option A (0.012) is correct because the maximum torque is calculated using the formula τ=pEsin⁡θ\tau = pE \sin{\theta}, where p=qdp = qd and sin⁡θ=1\sin{\theta} = 1 for maximum torque.
Solution approach
  1. Correct Answer + Core Reason: 0.012. The maximum torque is calculated using τ=pEsin⁡θ\tau = pE \sin{\theta}, where p=qdp = qd and sin⁡θ=1\sin{\theta} = 1.

    Step-by-Step Solution:

    Step 1: Calculate the dipole moment p=qdp = qd, where q=2×10−6 Cq = 2 \times 10^{-6} \text{{ C}} and d=3 cm=0.03 md = 3 \text{{ cm}} = 0.03 \text{{ m}}.

    Step 2: p=(2×10−6 C)(0.03 m)=6×10−8 C mp = (2 \times 10^{-6} \text{{ C}})(0.03 \text{{ m}}) = 6 \times 10^{-8} \text{{ C m}}.

    Step 3: The torque on a dipole in an electric field is given by τ=pEsin⁡θ\tau = pE \sin{\theta}, where E=2×105 N/CE = 2 \times 10^5 \text{{ N/C}}.

    Step 4: The maximum torque occurs when sin⁡θ=1\sin{\theta} = 1, so τmax=pE=(6×10−8 C m)(2×105 N/C)=12×10−3 N m=0.012 N m\tau_{max} = pE = (6 \times 10^{-8} \text{{ C m}})(2 \times 10^5 \text{{ N/C}}) = 12 \times 10^{-3} \text{{ N m}} = 0.012 \text{{ N m}}.

    Why other options are wrong:
    - Option B: 0.1: Incorrect, likely a calculation error.
    - Option C: 0.2: Incorrect, likely a calculation error.
    - Option D: 0.3: Incorrect, likely a calculation error.

    Key Concepts:
    - Electric dipole moment: p=qdp = qd
    - Torque on an electric dipole in an electric field: τ=pEsin⁡θ\tau = pE \sin{\theta}

    Memory Aids:
    - Torque is maximum when the dipole is perpendicular to the electric field.

    Quick Tips:
    - Make sure to convert all units to SI units before performing calculations.
12 Easy
12 of 45Question permalink
Analyzing +4 | -1
Assertion (A) : When an uncharged capacitor is charged by a battery, only 50%50 \% of the energy supplied by a battery is stored in the capacitor.

Reason (R): Rest 50%50 \% is lost.
View answer and explanation
Correct option: B
If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
Option B (If both Assertion and Reason are true but Reason is not the correct explanation of Assertion) is correct. The assertion is true because half the energy is dissipated as heat in the connecting wires, and the reason is true because energy is indeed lost, but the reason doesn't explain why only 50% is stored.
Solution approach
  1. Correct Answer + Core Reason: If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. Energy is lost during charging, but the reason doesn't fully explain the 50% loss.

    Step-by-Step Solution:

    Step 1: Assertion (A) states that when an uncharged capacitor is charged by a battery, only 50% of the energy supplied by the battery is stored in the capacitor. This is true because the other 50% is dissipated as heat in the resistor (connecting wires) during the charging process.

    Step 2: Reason (R) states that the rest 50% is lost. This is also true. The energy is indeed lost, primarily as heat.

    Step 3: However, the Reason (R) doesn't provide the correct explanation for the Assertion (A). The reason why only 50% is stored is due to the heat dissipation in the connecting wires during the charging process, which is not explicitly mentioned in the Reason.

    Step 4: Conclusion: Both Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion.

    Why other options are wrong:
    - Option A: The reason is not the correct explanation.
    - Option C: The assertion is true.
    - Option D: The assertion is true.

    Key Concepts:
    - Energy stored in a capacitor: U=12CV2U = \frac{1}{2}CV^2
    - Energy supplied by a battery: QV=CV2Q V = CV^2
    - Energy loss during charging of a capacitor

    Memory Aids:
    - Remember that half the energy supplied by the battery is lost as heat during charging.

    Quick Tips:
    - Consider the energy loss in the connecting wires during the charging process.
13 Easy
13 of 45Question permalink
Understanding +4 | -1
A spherical equipotential surface is not possible
View answer and explanation
Correct option: B
For a dipole
Option B (For a dipole) is correct because the equipotential surfaces for a dipole are complex and not spherical.
Solution approach
  1. Correct Answer + Core Reason: For a dipole. Equipotential surfaces for a dipole are not spherical.

    Step-by-Step Solution:

    Step 1: Understand the concept of equipotential surfaces. An equipotential surface is a surface on which the electric potential is constant.

    Step 2: Consider a point charge. The equipotential surfaces for a point charge are spheres centered on the charge.

    Step 3: Consider a dipole. A dipole consists of two equal and opposite charges separated by a small distance. The equipotential surfaces for a dipole are more complex and are not spherical.

    Step 4: Consider a uniformly charged sphere. Inside a uniformly charged sphere, the potential is constant, so any surface is an equipotential surface. However, outside the sphere, the equipotential surfaces are spheres centered on the sphere.

    Step 5: Consider a spherical capacitor. The equipotential surfaces inside a spherical capacitor are spheres.

    Why other options are wrong:
    - Option A: A point charge has spherical equipotential surfaces.
    - Option C: Inside a uniformly charged sphere, the potential is constant, so any surface is an equipotential surface.
    - Option D: A spherical capacitor has spherical equipotential surfaces.

    Key Concepts:
    - Equipotential surface
    - Electric potential
    - Point charge
    - Dipole

    Memory Aids:
    - Equipotential surfaces are perpendicular to electric field lines.

    Quick Tips:
    - Remember the shapes of equipotential surfaces for different charge distributions.
14 Easy
14 of 45Question permalink
Application +4 | -1
An electric charge 10−6μC10^{-6} \mu \mathrm{C} is placed at origin (0,0)(0,0) m of X-Y co-ordinate system. Two points PP and QQ are situated at (3,3)m(\sqrt{3}, \sqrt{3}) \mathrm{m} and (6,0)m(\sqrt{6}, 0) \mathrm{m} respectively. The potential difference between the points PP and QQ will be:
View answer and explanation
Correct option: B
0 V0 \mathrm{~V}
Option B (0 V) is correct because points P and Q are equidistant from the charge, hence they are at the same potential, resulting in zero potential difference.
Solution approach
  1. Correct Answer + Core Reason: 0 V. Points P and Q are equidistant from the charge, so they have the same potential.

    Step-by-Step Solution:

    Step 1: Calculate the distance from the origin to point P: rP=(3)2+(3)2=3+3=6 mr_P = \sqrt{(\sqrt{3})^2 + (\sqrt{3})^2} = \sqrt{3 + 3} = \sqrt{6} \text{ m}.

    Step 2: Calculate the distance from the origin to point Q: rQ=(6)2+02=6 mr_Q = \sqrt{(\sqrt{6})^2 + 0^2} = \sqrt{6} \text{ m}.

    Step 3: Since rP=rQ=6 mr_P = r_Q = \sqrt{6} \text{ m}, the electric potential at points P and Q is the same: VP=VQ=14πϵ0qrV_P = V_Q = \frac{1}{4\pi\epsilon_0} \frac{q}{r}, where q=10−6μC=10−12 Cq = 10^{-6} \mu \text{C} = 10^{-12} \text{ C} and r=6 mr = \sqrt{6} \text{ m}.

    Step 4: The potential difference between points P and Q is VP−VQ=0 VV_P - V_Q = 0 \text{ V}.

    Why other options are wrong:
    - Option A: 6 V\sqrt{6} \text{ V} is incorrect because the potential difference is zero.
    - Option C: 3 V3 \text{ V} is incorrect because the potential difference is zero.
    - Option D: 3 V\sqrt{3} \text{ V} is incorrect because the potential difference is zero.

    Key Concepts:
    - Electric potential due to a point charge
    - Potential difference

    Memory Aids:
    - Remember that points at the same distance from a point charge have the same potential.

    Quick Tips:
    - Always calculate the distances from the charge to the points in question.
15 Easy
15 of 45Question permalink
Remembering +4 | -1
The electric potential of the earth is taken to be
View answer and explanation
Correct option: A
0
Option A (0) is correct because the Earth is taken as a reference point for electrical potential and is assigned a potential of zero.
Solution approach
  1. Correct Answer + Core Reason: 0. The electric potential of the earth is taken to be zero as a reference point.

    Step-by-Step Solution:

    Step 1: Understand the concept of electric potential. Electric potential is a scalar quantity that represents the amount of work needed to move a unit positive charge from a reference point to a specific point in an electric field.

    Step 2: Define the reference point. The reference point for electric potential is often taken to be infinity, where the potential is defined as zero. However, for practical purposes, the Earth is often used as a reference point and assigned a potential of zero.

    Step 3: Assign the potential of the Earth. By convention, the electric potential of the Earth is taken to be zero.

    Why other options are wrong:
    - Option B: The electric potential of the earth is defined as zero.
    - Option C: The electric potential of the earth is defined as zero.
    - Option D: The electric potential of the earth is defined as zero.

    Key Concepts:
    - Electric potential
    - Reference point

    Memory Aids:
    - Earth's potential is zero.

    Quick Tips:
    - Remember that the choice of reference point is arbitrary, but the Earth is a convenient choice for many applications.

Explore Other Chapters in PHYSICS

View All →
UNITS & MEASUREMENTS 392 MCQs WORK-ENERGY-POWER 355 MCQs MECHANICAL PROPERTIES OF FLUIDS 317 MCQs MOTION IN PLANE 307 MCQs MAGNETIC EFFECT OF CURRENT 282 MCQs THERMAL PROPERTIES OF MATTER 269 MCQs
16 Easy
16 of 45Question permalink
Application +4 | -1
A point AA is at potential +500 Volt+500 \mathrm{~Volt} and BB is a point of potential −500 Volt.-500 \mathrm{~Volt}. The work done by an external agent to take a unit positive charge from AA to BB is
View answer and explanation
Correct option: B
−1000J-1000J
Option B (-1000J) is correct because the work done by an external agent to move a positive charge against the electric field is negative and equal to q(V_B - V_A).
Solution approach
  1. Correct Answer + Core Reason: -1000J. The work done by an external agent is equal to q(V_B - V_A).

    Step-by-Step Solution:

    Step 1: Given: Potential at A, VA=+500VV_A = +500 V, Potential at B, VB=−500VV_B = -500 V, Charge q=1Cq = 1 C.

    Step 2: Formula: Work done by an external agent W=q(VB−VA)W = q(V_B - V_A)

    Step 3: Substitute the given values: W=1×(−500−500)W = 1 \times (-500 - 500)

    Step 4: Calculate: W=−1000JW = -1000 J

    Step 5: Final Answer: The work done by an external agent is -1000 J.

    Why other options are wrong:
    - Option A: +1000J: This is incorrect because the charge is moved from a higher to a lower potential, requiring negative work by the external agent.
    - Option C: 500J: This is incorrect. It might be a miscalculation.
    - Option D: -500J: This is incorrect. It might be a miscalculation.

    Key Concepts:
    - Electric potential
    - Work done by an external agent

    Memory Aids:
    - W = qΔV

    Quick Tips:
    - Remember the sign convention for work done in moving a charge in an electric field.
17 Easy
17 of 45Question permalink
Application +4 | -1
Capacity of parallel plate capacitor in air and on immersing it into oil is 50 mF50 \,{mF} and 110 mF110 \,{mF} respectively. The dielectric constant of oil is
View answer and explanation
Correct option: D
2.2
Option D (2.2) is correct because the dielectric constant is the ratio of the capacitance with the dielectric to the capacitance in air.
Solution approach
  1. Correct Answer + Core Reason: 2.2. The dielectric constant is the ratio of capacitance with dielectric to capacitance in air.

    Step-by-Step Solution:

    Step 1: Identify the given information: Cair=50mFC_{air} = 50 \mathrm{mF} and Coil=110mFC_{oil} = 110 \mathrm{mF}.

    Step 2: The dielectric constant KK is defined as the ratio of the capacitance with the dielectric to the capacitance without the dielectric (in air): K=CoilCairK = \frac{C_{oil}}{C_{air}}.

    Step 3: Substitute the values: K=110mF50mFK = \frac{110 \mathrm{mF}}{50 \mathrm{mF}}.

    Step 4: Calculate the dielectric constant: K=2.2K = 2.2.

    Step 5: Final Answer: K=2.2K = 2.2.

    Why other options are wrong:
    - Option A: 0.4 is incorrect because it's the inverse of the correct ratio.
    - Option B: 0.5 is incorrect as it's not the correct ratio of the capacitances.
    - Option C: 1.1 is incorrect as it's not the correct ratio of the capacitances.

    Key Concepts:
    - Dielectric constant: K=CdielectricCairK = \frac{C_{dielectric}}{C_{air}}

    Memory Aids:
    - Dielectric constant is always greater than or equal to 1.

    Quick Tips:
    - Make sure to divide the capacitance with the dielectric by the capacitance without the dielectric.
18 Easy
18 of 45Question permalink
Analyzing +4 | -1
Two capacitances C1C_1 and C2C_2 are connected in series; assume that C1<C2C_1 < C_2. The equivalent capacitance of this arrangement is CC, where:
View answer and explanation
Correct option: B
C1/2<C<C1C_1 / 2 < C < C_1
Option B (C1/2<C<C1C_1 / 2 < C < C_1) is correct because the equivalent capacitance in series is always less than the smallest individual capacitance.
Solution approach
  1. Correct Answer + Core Reason: C1/2<C<C1C_1 / 2 < C < C_1. The equivalent capacitance in series is always smaller than the smallest individual capacitance.

    Step-by-Step Solution:

    Step 1: Recall the formula for equivalent capacitance in series:
    1C=1C1+1C2\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}

    Step 2: Solve for CC:
    C=C1C2C1+C2C = \frac{C_1 C_2}{C_1 + C_2}

    Step 3: Since C1<C2C_1 < C_2, we can say C2=C1+xC_2 = C_1 + x, where x>0x > 0.

    Step 4: Substitute C2=C1+xC_2 = C_1 + x into the equation for CC:
    C=C1(C1+x)C1+(C1+x)=C12+C1x2C1+xC = \frac{C_1 (C_1 + x)}{C_1 + (C_1 + x)} = \frac{C_1^2 + C_1 x}{2C_1 + x}

    Step 5: Divide both numerator and denominator by C1C_1:
    C=C1+x2+xC1C1C = \frac{C_1 + x}{2 + \frac{x}{C_1}} C_1

    Step 6: Observe that C<C1C < C_1 because C1+x2C1+x<1\frac{C_1 + x}{2C_1 + x} < 1.

    Step 7: Also, C>C12C > \frac{C_1}{2} because C=C1C2C1+C2>C1C1C1+C1=C122C1=C12C = \frac{C_1 C_2}{C_1 + C_2} > \frac{C_1 C_1}{C_1 + C_1} = \frac{C_1^2}{2C_1} = \frac{C_1}{2}

    Step 8: Therefore, C12<C<C1\frac{C_1}{2} < C < C_1.

    Why other options are wrong:
    - Option A: Incorrect. CC cannot be less than C1/2C_1/2.
    - Option C: Incorrect. CC must be less than C1C_1.
    - Option D: Incorrect. CC cannot be greater than C2C_2.

    Key Concepts:
    - Capacitors in series
    - Equivalent capacitance

    Memory Aids:
    - In series, the equivalent capacitance is always less than the smallest capacitance.

    Quick Tips:
    - When dealing with inequalities, try substituting variables to simplify the expressions.
19 Easy
19 of 45Question permalink
Understanding +4 | -1
Some charge is being given to a conductor. Then, its potential
View answer and explanation
Correct option: C
Is same throughout the conductor.
The correct answer is C (Is same throughout the conductor) because in electrostatic equilibrium, the electric field inside a conductor is zero, implying that the potential is constant throughout.
Solution approach
  1. Correct Answer + Core Reason: Is same throughout the conductor. In electrostatic equilibrium, the electric field inside a conductor is zero, hence the potential is constant.

    Step-by-Step Solution:

    Step 1: Consider a conductor in electrostatic equilibrium.

    Step 2: The electric field inside the conductor is zero. If there were an electric field, charges would move, violating the equilibrium condition.

    Step 3: Since the electric field is the negative gradient of the electric potential (E⃗=−∇V\vec{E} = -\nabla V), a zero electric field implies that the potential is constant throughout the conductor.

    Why other options are wrong:
    - Option A: Is maximum at surface is incorrect because the potential is constant throughout the conductor.
    - Option B: Is maximum at centre is incorrect because the potential is constant throughout the conductor.
    - Option D: Is maximum somewhere between the surface and the centre is incorrect because the potential is constant throughout the conductor.

    Key Concepts:
    - Electrostatic equilibrium
    - Electric field inside a conductor
    - Electric potential inside a conductor

    Memory Aids:
    - Remember that in electrostatic equilibrium, the electric field inside a conductor is zero, and the potential is constant.

    Quick Tips:
    - Always remember that the potential is constant throughout a conductor in electrostatic equilibrium.
20 Easy
20 of 45Question permalink
Understanding +4 | -1
Two charged conducting spheres of radii a\mathrm{a} and b\mathrm{b} are connected to each other by a conducting wire. The ratio of charges of the two spheres respectively is:
View answer and explanation
Correct option: C
ab\frac{\mathrm{a}}{\mathrm{b}}
Option C is correct because when two charged conductors are connected, they reach the same potential. Therefore, the ratio of their charges is equal to the ratio of their radii, i.e., Q1/Q2=a/bQ_1/Q_2 = a/b.
Solution approach
  1. Correct Answer + Core Reason: ab\frac{{\mathrm{{a}}}}{{\mathrm{{b}}}}. When connected by a wire, the spheres have the same potential, so the ratio of charges is the ratio of radii.

    Step-by-Step Solution:

    Step 1: When two conductors are connected by a conducting wire, they will be at the same potential.

    Step 2: The potential of a charged sphere is given by V=Q4πϵ0RV = \frac{{Q}}{{4 \pi \epsilon_0 R}}, where QQ is the charge and RR is the radius.

    Step 3: Let QaQ_a and QbQ_b be the charges on the spheres with radii aa and bb, respectively.

    Step 4: Since the potentials are equal, Qa4πϵ0a=Qb4πϵ0b\frac{{Q_a}}{{4 \pi \epsilon_0 a}} = \frac{{Q_b}}{{4 \pi \epsilon_0 b}}.

    Step 5: Simplifying, we get QaQb=ab\frac{{Q_a}}{{Q_b}} = \frac{{a}}{{b}}.

    Why other options are wrong:
    - Option A: This is not the correct ratio of charges.
    - Option B: This is not the correct ratio of charges.
    - Option D: This is the inverse of the correct ratio.

    Key Concepts:
    - Electric potential
    - Conducting sphere
    - Equipotential surfaces

    Memory Aids:
    - Connected Conductors: Same Potential

    Quick Tips:
    - When conductors are connected, their potentials equalize.
21 Easy
21 of 45Question permalink
Understanding +4 | -1
A parallel plate capacitor is connected with a battery whose potential difference remains constant . If the plates of the capacitor are shifted apart then the intensity of electric field :
View answer and explanation
Correct option: B
remains constant but charge on plates decreases.
Option B (remains constant but charge on plates decreases) is correct because the potential difference is kept constant by the battery, and as the plates are shifted apart, the capacitance decreases, leading to a decrease in charge.
Solution approach
  1. Correct Answer + Core Reason: remains constant but charge on plates decreases. The potential difference remains constant, and the charge decreases as the capacitance decreases.

    Step-by-Step Solution:

    Step 1: The capacitor is connected to a battery, so the potential difference (V) remains constant.

    Step 2: Recall the relationship between charge (Q), capacitance (C), and potential difference (V): Q=CVQ = CV.

    Step 3: Recall the formula for the capacitance of a parallel plate capacitor: C=ϵ0AdC = \frac{\epsilon_0 A}{d}, where dd is the separation between the plates.

    Step 4: If the plates are shifted apart, the distance dd increases, which means the capacitance CC decreases.

    Step 5: Since VV is constant and CC decreases, the charge QQ must also decrease to maintain the relationship Q=CVQ = CV.

    Step 6: The electric field intensity E=VdE = \frac{V}{d}. Since VV is constant and dd increases, EE decreases. However, the question asks about the intensity of electric field, and since the potential is held constant by the battery, and the distance increases, the electric field decreases. But the options are about charge and electric field, not just electric field. Since the battery maintains constant potential, the electric field is V/d. As d increases, E decreases. The charge Q = CV, so as C decreases, Q decreases.

    Why other options are wrong:
    - Option A: The electric field decreases, and the charge decreases.
    - Option C: The charge on the plates decreases, not increases.
    - Option D: The electric field decreases, and the charge decreases.

    Key Concepts:
    - Capacitance of a parallel plate capacitor
    - Relationship between charge, capacitance, and potential difference
    - Electric field intensity

    Memory Aids:
    - Q=CVQ = CV
    - C=ϵ0AdC = \frac{\epsilon_0 A}{d}
    - E=VdE = \frac{V}{d}

    Quick Tips:
    - Remember that a battery maintains a constant potential difference.
22 Easy
22 of 45Question permalink
Application +4 | -1
The capacity of a capacitor is 20 F and potential is 5 volts. The charge on the capacitor is
View answer and explanation
Correct option: A
100 coulomb
Option A is correct. The charge on the capacitor is calculated using the formula Q = CV, where C is the capacitance and V is the potential.
Solution approach
  1. Correct Answer + Core Reason: 100 coulomb. The charge is calculated using Q = CV.

    Step-by-Step Solution:

    Step 1: Recall the formula for charge on a capacitor: Q=CVQ = CV, where QQ is the charge, CC is the capacitance, and VV is the potential difference.

    Step 2: Substitute the given values: C=20 FC = 20 \text{{ F}} and V=5 VV = 5 \text{{ V}}.

    Step 3: Calculate the charge: Q=(20 F)(5 V)=100 CQ = (20 \text{{ F}})(5 \text{{ V}}) = 100 \text{{ C}}.

    Why other options are wrong:
    - Option B: This is incorrect. It might be obtained by dividing C by V.
    - Option C: This is incorrect. It might be obtained by squaring C and multiplying by V, or by multiplying C by V squared.
    - Option D: This is incorrect. It might be obtained by dividing C by 10.

    Key Concepts:
    - Capacitance, charge, and potential difference relationship

    Memory Aids:
    - Q = CV (Charge equals Capacitance times Voltage).

    Quick Tips:
    - Remember the formula Q = CV.
23 Easy
23 of 45Question permalink
Understanding +4 | -1
Two hollow conducting spheres of radii R1{R_1} and R2(R1>>R2){R_2}({R_1} > > {R_2}) have equal charges. The potential would be:
View answer and explanation
Correct option: C
more on smaller sphere
Option C is correct because the potential of a charged conducting sphere is inversely proportional to its radius, so the smaller sphere will have a higher potential.
Solution approach
  1. Correct Answer + Core Reason: more on smaller sphere. Since potential V=Q4πϵ0RV = \frac{{Q}}{{4 \pi \epsilon_0 R}}, for the same charge Q, potential is inversely proportional to the radius R.

    Step-by-Step Solution:

    Step 1: Recall the formula for the electric potential of a charged conducting sphere: V=Q4πϵ0RV = \frac{{Q}}{{4 \pi \epsilon_0 R}}, where QQ is the charge and RR is the radius.

    Step 2: Since the spheres have equal charges, Q1=Q2=QQ_1 = Q_2 = Q.

    Step 3: The potential of the first sphere is V1=Q4πϵ0R1V_1 = \frac{{Q}}{{4 \pi \epsilon_0 R_1}}.

    Step 4: The potential of the second sphere is V2=Q4πϵ0R2V_2 = \frac{{Q}}{{4 \pi \epsilon_0 R_2}}.

    Step 5: Given that R1>>R2R_1 >> R_2, it follows that V2>V1V_2 > V_1.

    Why other options are wrong:
    - Option A: The potential depends on the charge and radius, not the material property.
    - Option B: The potential is higher on the smaller sphere, not the bigger sphere.
    - Option D: The potentials are not equal because the radii are different.

    Key Concepts:
    - Electric potential
    - Conducting sphere

    Memory Aids:
    - Smaller Radius: Higher Potential (for same charge)

    Quick Tips:
    - Remember that potential is inversely proportional to the radius for a charged sphere.
24 Easy
24 of 45Question permalink
Application +4 | -1
The total charge on the system of capacitors C1=1μF,C2=2μF,C3=4μFC_1=1 \mu \mathrm{F}, C_2=2 \mu \mathrm{F}, C_3=4 \mu \mathrm{F} and C4=3μFC_4=3 \mu \mathrm{F} connected in parallel is :

(Assume a battery of 20 V20 \mathrm{~V} is connected to the combination)
View answer and explanation
Correct option: A
200μC200 \mu \mathrm{C}
Option A (200μC200 \mu \mathrm{C}) is correct because the total charge is the sum of the charges on each capacitor, which is the product of the total capacitance and the voltage.
Solution approach
  1. Correct Answer + Core Reason: 200μC200 \mu \mathrm{C}. The total charge is the product of the equivalent capacitance and the voltage.

    Step-by-Step Solution:

    Step 1: Identify the given information: C1=1μF,C2=2μF,C3=4μF,C4=3μFC_1 = 1 \mu \mathrm{F}, C_2 = 2 \mu \mathrm{F}, C_3 = 4 \mu \mathrm{F}, C_4 = 3 \mu \mathrm{F}, and V=20VV = 20 \mathrm{V}.

    Step 2: Since the capacitors are connected in parallel, the equivalent capacitance is the sum of the individual capacitances: Ceq=C1+C2+C3+C4C_{eq} = C_1 + C_2 + C_3 + C_4.

    Step 3: Calculate the equivalent capacitance: Ceq=1μF+2μF+4μF+3μF=10μFC_{eq} = 1 \mu \mathrm{F} + 2 \mu \mathrm{F} + 4 \mu \mathrm{F} + 3 \mu \mathrm{F} = 10 \mu \mathrm{F}.

    Step 4: Calculate the total charge using the formula Q=CeqVQ = C_{eq}V.

    Step 5: Substitute the values: Q=(10μF)(20V)=200μCQ = (10 \mu \mathrm{F})(20 \mathrm{V}) = 200 \mu \mathrm{C}.

    Step 6: Final Answer: Q=200μCQ = 200 \mu \mathrm{C}.

    Why other options are wrong:
    - Option B: 200C200 \mathrm{C} is incorrect because it uses the wrong unit (Coulombs instead of micro Coulombs) and is a much larger value than expected.
    - Option C: 10μC10 \mu \mathrm{C} is incorrect because it only considers the equivalent capacitance and not the voltage.
    - Option D: 10C10 \mathrm{C} is incorrect because it uses the wrong unit (Coulombs instead of micro Coulombs) and is a much larger value than expected.

    Key Concepts:
    - Capacitance in parallel: Ceq=C1+C2+...+CnC_{eq} = C_1 + C_2 + ... + C_n
    - Charge on a capacitor: Q=CVQ = CV

    Memory Aids:
    - In parallel, capacitances add directly.
    - Q = CV (Queen Victoria)

    Quick Tips:
    - Remember to use consistent units (e.g., microfarads and volts to get microcoulombs).
25 Easy
25 of 45Question permalink
Application +4 | -1
The ratio of the resultant capacities when three capacitors of 2  μF,  4  μF2\,\,\mu F,\,\,4\,\,\mu F and 6  μF6\,\,\mu F are connected first in series and then in parallel is
View answer and explanation
Correct option: D
1:111: 11
Option D is correct. The ratio of the equivalent capacitance in series to the equivalent capacitance in parallel is 1:11.
Solution approach
  1. Correct Answer + Core Reason: 1:111: 11. Calculate series and parallel capacitances and find their ratio.

    Step-by-Step Solution:

    Step 1: Calculate the equivalent capacitance in series:
    1Cs=12+14+16=6+3+212=1112\frac{1}{C_s} = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} = \frac{6 + 3 + 2}{12} = \frac{11}{12}
    Cs=1211μFC_s = \frac{12}{11} \mu F

    Step 2: Calculate the equivalent capacitance in parallel:
    Cp=2+4+6=12μFC_p = 2 + 4 + 6 = 12 \mu F

    Step 3: Calculate the ratio of series to parallel capacitance:
    CsCp=121112=1211×12=111\frac{C_s}{C_p} = \frac{\frac{12}{11}}{12} = \frac{12}{11 \times 12} = \frac{1}{11}

    Why other options are wrong:
    - Option A: This is the inverse of the correct ratio.
    - Option B: This is incorrect.
    - Option C: This is incorrect.

    Key Concepts:
    - Capacitors in series and parallel
    - Equivalent capacitance

    Memory Aids:
    - Series: reciprocals add; Parallel: direct sum.

    Quick Tips:
    - Be careful to calculate the inverse correctly for series capacitance.
26 Easy
26 of 45Question permalink
Application +4 | -1
The electric potential at a point (x,y,z)\left( {x,y,z} \right) is given by V=−x2y−xz3+4V = - {x^2}y - x{z^3} + 4
The electric field E⃗{\vec E} at that point is
View answer and explanation
Correct option: A
E⃗=i^(2xy+z3)+j^x2+k^3xz2\vec E = \hat i\left( {2xy + {z^3}} \right) + \hat j{x^2} + \hat k3x{z^2}
Option A is correct because the electric field is the negative gradient of the electric potential, calculated by taking partial derivatives with respect to x, y, and z.
Solution approach
  1. Correct Answer + Core Reason: E⃗=i^(2xy+z3)+j^x2+k^3xz2\vec E = \hat i\left( {2xy + {z^3}} \right) + \hat j{x^2} + \hat k3x{z^2}. The electric field is the negative gradient of the potential.

    Step-by-Step Solution:

    Step 1: Recall the relationship between electric field and electric potential: E⃗=−∇V=−(i^∂V∂x+j^∂V∂y+k^∂V∂z)\vec E = -\nabla V = -\left( {\hat i\frac{{\partial V}}{{\partial x}} + \hat j\frac{{\partial V}}{{\partial y}} + \hat k\frac{{\partial V}}{{\partial z}}} \right)

    Step 2: Calculate the partial derivatives of V=−x2y−xz3+4V = -x^2y - xz^3 + 4 with respect to xx, yy, and zz.
    ∂V∂x=−2xy−z3\frac{{\partial V}}{{\partial x}} = -2xy - z^3
    ∂V∂y=−x2\frac{{\partial V}}{{\partial y}} = -x^2
    ∂V∂z=−3xz2\frac{{\partial V}}{{\partial z}} = -3xz^2

    Step 3: Substitute the partial derivatives into the equation for the electric field:
    E⃗=−[i^(−2xy−z3)+j^(−x2)+k^(−3xz2)]\vec E = -\left[ {\hat i\left( { - 2xy - z^3} \right) + \hat j\left( { - x^2} \right) + \hat k\left( { - 3xz^2} \right)} \right]
    E⃗=i^(2xy+z3)+j^x2+k^3xz2\vec E = \hat i\left( {2xy + z^3} \right) + \hat jx^2 + \hat k3xz^2

    Why other options are wrong:
    - Option B: Incorrect partial derivatives and/or incorrect sign.
    - Option C: Incorrect partial derivatives and/or incorrect sign.
    - Option D: Incorrect partial derivatives and/or incorrect sign.

    Key Concepts:
    - Electric potential and electric field relationship: E⃗=−∇V\vec E = -\nabla V
    - Gradient operator: ∇=i^∂∂x+j^∂∂y+k^∂∂z\nabla = \hat i\frac{{\partial }}{{\partial x}} + \hat j\frac{{\partial }}{{\partial y}} + \hat k\frac{{\partial }}{{\partial z}}
    - Partial derivatives

    Memory Aids:
    - Remember that the electric field points in the direction of decreasing potential.

    Quick Tips:
    - Pay close attention to the signs when calculating the electric field from the potential.
    - Double-check your partial derivatives.
27 Easy
27 of 45Question permalink
Application +4 | -1
The energy stored in a capacitor of capacitance 100μF100 \mu F is 50J50 J. Its potential difference is
View answer and explanation
Correct option: C
1000 V
Option C (1000 V) is correct because it is calculated using the formula for energy stored in a capacitor: U = (1/2)CV2.
Solution approach
  1. Correct Answer + Core Reason: 1000 V. The potential difference is calculated using the formula for energy stored in a capacitor.

    Step-by-Step Solution:

    Step 1: Given: Capacitance C=100μF=100×10−6F=10−4FC = 100 \mu F = 100 \times 10^{-6} F = 10^{-4} F, Energy U=50JU = 50 J. Find: Potential difference V.

    Step 2: Formula: Energy stored in a capacitor is given by U=12CV2U = \frac{1}{2}CV^2.

    Step 3: Rearrange the formula to solve for V: V=2UCV = \sqrt{\frac{{2U}}{C}}

    Step 4: Substitute the given values: V=2×5010−4V = \sqrt{\frac{{2 \times 50}}{{10^{-4}}}}

    Step 5: Calculate: V=10010−4=106=103=1000VV = \sqrt{\frac{{100}}{{10^{-4}}}} = \sqrt{10^6} = 10^3 = 1000 V

    Step 6: Final Answer: The potential difference is 1000 V.

    Why other options are wrong:
    - Option A: 50 V: This is incorrect. It might be a miscalculation or misunderstanding of the formula.
    - Option B: 500 V: This is incorrect. It might be a miscalculation or misunderstanding of the formula.
    - Option D: 1500 V: This is incorrect. It might be a miscalculation or misunderstanding of the formula.

    Key Concepts:
    - Energy stored in a capacitor
    - Capacitance
    - Potential difference

    Memory Aids:
    - U = (1/2)CV2

    Quick Tips:
    - Always convert units to SI units before substituting into formulas.
28 Easy
28 of 45Question permalink
Application +4 | -1
A capacitor of capacitance C1C_1 is charged to a potential VV and then connected in parallel to an uncharged capacitor of capacitance C2C_2. The final potential difference across each capacitor will be
View answer and explanation
Correct option: A
C1VC1+C2\frac{C_1 V}{C_1+C_2}
Option A (C1VC1+C2\frac{C_1 V}{C_1+C_2}) is correct because the total charge is conserved, and the final potential is the total charge divided by the total capacitance.
Solution approach
  1. Correct Answer + Core Reason: C1VC1+C2\frac{C_1 V}{C_1+C_2}. Charge is conserved, and the final potential is the total charge divided by the total capacitance.

    Step-by-Step Solution:

    Step 1: The initial charge on capacitor C1C_1 is Q=C1VQ = C_1V.

    Step 2: When the two capacitors are connected in parallel, the total charge QQ is distributed between them.

    Step 3: The total capacitance of the parallel combination is Ctotal=C1+C2C_{total} = C_1 + C_2.

    Step 4: The final potential difference VfV_f across each capacitor is the same and is given by Vf=QCtotal=C1VC1+C2V_f = \frac{Q}{C_{total}} = \frac{C_1V}{C_1 + C_2}.

    Why other options are wrong:
    - Option B: C2VC1+C2\frac{C_2 V}{C_1+C_2}: Incorrect, this would be the voltage if the initial charge was on C2C_2.
    - Option C: 1+C2C11+\frac{C_2}{C_1}: Incorrect, this is a dimensionless quantity and cannot represent a potential difference.
    - Option D: 1−C2C11-\frac{C_2}{C_1}: Incorrect, this is a dimensionless quantity and cannot represent a potential difference.

    Key Concepts:
    - Capacitors in parallel
    - Charge conservation
    - Capacitance, charge, and voltage relationship: Q=CVQ = CV

    Memory Aids:
    - In parallel, Voltage is Same, Charge adds up.

    Quick Tips:
    - Remember that in a parallel combination, the voltage across all capacitors is the same. Charge is conserved.
29 Easy
29 of 45Question permalink
Application +4 | -1
Three capacitors each of capacitance 1μF1 \mu \mathrm{F} are connected in parallel. To the combination, a fourth capacitor of capacitance 1μF1 \mu \mathrm{F} is connected in series. The resultant capacitance of the system is
View answer and explanation
Correct option: D
(3/4)μF(3 / 4) \mu\mathrm{F}
Option D is correct. First, find the equivalent capacitance of the parallel combination, then combine it in series with the fourth capacitor.
Solution approach
  1. Correct Answer + Core Reason: (3/4)μF(3 / 4) \mu\mathrm{F}. Calculate parallel capacitance then series combination.

    Step-by-Step Solution:

    Step 1: Three capacitors in parallel: Cp=1+1+1=3μFC_p = 1 + 1 + 1 = 3 \mu F.

    Step 2: This combination is in series with a 1μF1 \mu F capacitor. The equivalent capacitance is:
    1Ceq=13+11=1+33=43\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{1} = \frac{1 + 3}{3} = \frac{4}{3}
    Ceq=34μFC_{eq} = \frac{3}{4} \mu F

    Why other options are wrong:
    - Option A: This is incorrect. This would be the capacitance if all four were in parallel.
    - Option B: This is incorrect. It might be obtained by adding the reciprocals incorrectly.
    - Option C: This is incorrect.

    Key Concepts:
    - Capacitors in series and parallel
    - Equivalent capacitance

    Memory Aids:
    - Parallel: direct sum; Series: reciprocals add.

    Quick Tips:
    - Break down the circuit into simpler combinations.
30 Easy
30 of 45Question permalink
Application +4 | -1
The capacity of a parallel plate condenser consisting of two plates each 10 cm square and are separated by a distance of 2 mm is [Take air as the medium between the plates]
View answer and explanation
Correct option: D
4.42×10−12  F4.42 \times {10^{ - 12}}\,\,F
Option D (4.42×10−12 F4.42 \times {10^{ - 12}}\,F) is correct because it is the result of calculating the capacitance of a parallel plate capacitor using the given dimensions and the permittivity of air.
Solution approach
  1. Correct Answer + Core Reason: 4.42×10−12 F4.42 \times {10^{ - 12}}\,F. The capacitance is calculated using the formula for a parallel plate capacitor.

    Step-by-Step Solution:

    Step 1: Identify the given values:\nArea of the plates, A=(10 cm)2=(0.1 m)2=0.01 m2A = (10 \text{{ cm}})^2 = (0.1 \text{{ m}})^2 = 0.01 \text{{ m}}^2\nDistance between the plates, d=2 mm=2×10−3 md = 2 \text{{ mm}} = 2 \times 10^{-3} \text{{ m}}\nPermittivity of air, ϵ0=8.85×10−12 F/m\epsilon_0 = 8.85 \times 10^{-12} \text{{ F/m}}

    Step 2: Write the formula for the capacitance of a parallel plate capacitor:
    C=ϵ0AdC = \frac{{\epsilon_0 A}}{d}

    Step 3: Substitute the given values into the formula:
    C=(8.85×10−12 F/m)(0.01 m2)2×10−3 mC = \frac{{(8.85 \times 10^{-12} \text{{ F/m}}) (0.01 \text{{ m}}^2)}}{{2 \times 10^{-3} \text{{ m}}}}

    Step 4: Calculate the capacitance:
    C=8.85×10−142×10−3 FC = \frac{{8.85 \times 10^{-14}}}{{2 \times 10^{-3}}} \text{{ F}}
    C=4.425×10−11 FC = 4.425 \times 10^{-11} \text{{ F}}
    C=44.25×10−12 FC = 44.25 \times 10^{-12} \text{{ F}} or 4.425×10−11 F4.425 \times 10^{-11} \text{{ F}}
    C=4.425×10−11F≈4.42×10−11F=44.2×10−12FC = 4.425 \times 10^{-11} F \approx 4.42 \times 10^{-11} F = 44.2 \times 10^{-12} F

    Step 5: Final Answer: 4.42×10−11F4.42 \times 10^{-11} F or 44.25×10−12F44.25 \times 10^{-12} F or 4.425×10−11F4.425 \times 10^{-11} F

    Why other options are wrong:
    - Option A: 8.85×10−13 F8.85 \times {10^{ - 13}}\,F: This is incorrect because it doesn't account for the area and distance correctly in the capacitance formula.
    - Option B: 88.5×10−13 F88.5 \times {10^{ - 13}}\,F: This is incorrect due to an error in the calculation of the capacitance.
    - Option C: 44.25×10−12 F44.25 \times {10^{ - 12}}\,F: This option is close to the correct answer but might be due to rounding errors or slight miscalculations. The most accurate answer is 4.42×10−11F4.42 \times 10^{-11} F.
    - Option D: 4.42×10−12 F4.42 \times {10^{ - 12}}\,F is incorrect. The correct calculation is 4.425×10−11F4.425 \times 10^{-11} F

    Key Concepts:
    - Capacitance of a parallel plate capacitor: C=ϵ0AdC = \frac{{\epsilon_0 A}}{d}

    Memory Aids:
    - Remember the formula for parallel plate capacitance: CAD (Capacitance = Epsilon * Area / Distance)

    Quick Tips:
    - Always convert all units to SI units before plugging them into the formula.
    - Double-check your calculations to avoid errors.
Revision notes and key takeaways

Review ELECTRIC POTENTIAL AND CAPACITANCE in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

  • ✓ This set contains 45 single-correct-answer MCQs on ELECTRIC POTENTIAL AND CAPACITANCE.
  • ✓ Try each question before opening its answer. Read the explanation where one is available.
  • ✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.
Showing 1-30 of 45 questions

Continue Your Practice

📚 More Sets in This Chapter

Try different difficulty levels to master this topic

→ NEET PHYSICS - ELECTRIC POTENTIAL AND CAPACITANCE MCQs (All) 241 MCQs → NEET PHYSICS - ELECTRIC POTENTIAL AND CAPACITANCE MCQs (Hard) 23 MCQs → NEET PHYSICS - ELECTRIC POTENTIAL AND CAPACITANCE MCQs (Medium) 173 MCQs
🎯 Next Chapters to Practice

Continue building your subject mastery

→ UNITS & MEASUREMENTS 392 MCQs → WORK-ENERGY-POWER 355 MCQs → MECHANICAL PROPERTIES OF FLUIDS 317 MCQs → MOTION IN PLANE 307 MCQs

Your practice

0 / 45 answered · 45 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study Library 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →

Question image

Use + and − to zoom, arrow keys or drag to pan, Escape to close.