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JEE • PHYSICS • ELECTRIC POTENTIAL AND CAPACITANCE

Electric potential and capacitance practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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121 Medium
121 of 372
Application +4 | -1
Three capacitors of 2μF,  3μF2\mu F,\;3\mu F and 6μF6\mu F are joined in series and the combination is charged by means of a 24  volt24\;volt battery. The potential difference between the plates of the 6μF6\mu F capacitor is
View answer and explanation
Correct option: A
4  volt4\;volt
Option A (4  volt4\;volt) is correct because the potential difference across the 6μF6\mu F capacitor is calculated using the total charge and the equivalent capacitance of the series combination.
122 Medium
122 of 372
Application +4 | -1
Three concentric spherical shells have radii a, b and c(a<b<c)\mathrm{c}(\mathrm{a} < \mathrm{b} < \mathrm{c}) and have surface charge densities σ,−σ\sigma,-\sigma and σ\sigma respectively. If VA,VBV_A, V_B and VC\mathrm{V}_{\mathrm{C}} denote the potentials of the three shells, then, for VA=VCV_A=V_C, we get -
View answer and explanation
Correct option: D
c=a+bc=a+b
Option D (c=a+bc=a+b) is correct because it satisfies the condition VA=VCV_A = V_C when the potentials are calculated for the three concentric shells.
123 Medium
123 of 372
Understanding +4 | -1
Two charges +q and –qq are kept apart. Then at any point on the right bisector of line joining the two charges
View answer and explanation
Correct option: B
The electric potential is zero
Option B is correct because the electric potential is a scalar quantity and the potentials due to +q and -q cancel out at any point on the right bisector.
124 Medium
124 of 372
Application +4 | -1
The displacement of a charge QQ in the electric field E⃗=e1i^+e2j^+e3k^\vec E = {e_1}\hat i + {e_2}\hat j + {e_3}\hat k is r⃗=ai^+bj^\vec r = a\hat i + b\hat j. The work done is
View answer and explanation
Correct option: A
Q(ae1+be2)Q\left( {a{e_1} + b{e_2}} \right)
Option A (Q(ae1+be2)Q\left( {a{e_1} + b{e_2}} \right)) is correct because the work done is the dot product of the force (Q*E) and the displacement vector.
125 Medium
125 of 372
Analyzing +4 | -1
Two spheres of radii rr and RR carry charges qq and QQ respectively. When they are connected by a wire, there will be no loss of energy of the system if
View answer and explanation
Correct option: B
qR=Qr{qR}={Qr}
The correct answer is B (qR=Qr{qR}={Qr}) because for no energy loss, the potentials of the two spheres must be equal after they are connected.
126 Medium
126 of 372
Application +4 | -1
The work done in bringing a 20 coulomb charge from point AA to point BB for distance 0.2 m is 2 joule. The potential difference between the two points will be (in volt)
View answer and explanation
Correct option: C
0.1
Option C (0.1) is correct because the potential difference is calculated as the work done per unit charge, V=WQV = \frac{W}{Q}.
127 Medium
127 of 372
Application +4 | -1
Three charges each 20μC20\mu C are placed at the corners of an equilateral triangle of side 0.4m0.4m. The potential energy of the systems is…… JJ
View answer and explanation
Correct option: A
27
Option A (27) is correct because it is the potential energy calculated using the formula for potential energy of a system of charges.
128 Medium
128 of 372
Application +4 | -1
Two square plates are at potential difference of 100 V separated by 2 cm. calculate electric intensity between them.
View answer and explanation
Correct option: B
5000
Option B (5000) is correct because the electric field intensity is calculated by dividing the potential difference by the distance between the plates.
129 Medium
129 of 372
Understanding +4 | -1
The electric field at a point on the equatorial plane at a distance r\text{r} from the centre of a dipole having dipole moment P\text{P} is given by (r≫(\text{r} \gg separation of two charges forming the dipole, ε0\varepsilon_0= permittivity of free space)
View answer and explanation
Correct option: D
E=−P4πε0r3\mathrm{E}=-\frac{\mathrm{P}}{4 \pi \varepsilon_0 \mathrm{r}^3}
Option D (E=−P4πε0r3\mathrm{E}=-\frac{\mathrm{P}}{4 \pi \varepsilon_0 \mathrm{r}^3}) is correct because it represents the electric field at a point on the equatorial plane of a dipole.
130 Medium
130 of 372
Application +4 | -1
A capacitor has capacitance 2 F2 \mathrm{~F}. plate separation 0.5 cm0.5 \mathrm{~cm} then area of plate
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the calculated area is 1.130×109m21.130 \times 10^9 m^2, which is not among the given options.
131 Medium
131 of 372
Application +4 | -1
On moving a charge of 30 coulombs by 3 cm, 6 J of work is done, then the potential difference between the points is
View answer and explanation
Correct option: A
0.2 V0.2 \mathrm{~V}
Option A (0.2 V0.2 \mathrm{~V}) is correct because the potential difference is the work done per unit charge, which is calculated as 6 J / 30 C.
132 Medium
132 of 372
Application +4 | -1
1000 small water drops each of radius rr and charge qq coalesce together to form one spherical drop. The potential of the big drop is larger than that of the smaller drop by a factor of
View answer and explanation
Correct option: B
100
Option B (100) is correct because the potential of a larger drop formed by coalescing smaller drops is proportional to n(2/3), where n is the number of smaller drops.
133 Medium
133 of 372
Understanding +4 | -1
An insulated charged sphere of radius 5 cm has
a potential of 10 V at the surface. The potential
at centre will be
View answer and explanation
Correct option: C
10 V
Option C is correct because the potential inside a charged sphere is constant and equal to the potential at the surface.
134 Medium
134 of 372
Application +4 | -1
If 20 J20 \mathrm{~J} of work has to be done to move an electric charge of 4C4 \mathrm{C} from a point, where potential is 10 V10 \mathrm{~V} to another point, where potential is V\mathrm{V}, find the value of V\mathrm{V}.
View answer and explanation
Correct option: D
15 V15 \mathrm{~V}
Option D (15 V) is correct because the potential difference multiplied by the charge equals the work done.
135 Medium
135 of 372
Application +4 | -1
A 4  μF4\,\,\mu F conductor is charged to 400  V\mathrm{400}\,\,\mathrm{V} and then its plates are joined through a resistance of 1  KΩ\mathrm{1}\,\,\mathrm{K\Omega }. The heat produced in the resistance is
View answer and explanation
Correct option: D
0.32  J\begin{array}{llllllllllllllllllll} {\mathrm{0}\mathrm{.32}\,\,\mathrm{J}} \end{array}
Option D (0.32 J) is correct because the energy stored in the capacitor is converted into heat in the resistor, and the energy stored is calculated by 12CV2\frac{1}{2}CV^2.
136 Medium
136 of 372
Application +4 | -1
A solid sphere of radius R\mathrm{R} is charged uniformly. At what distance from its surface is the electrostatic potential half of the potential at the centre
View answer and explanation
Correct option: C
R3\frac{R}{3}
Option C (R/3R/3) is correct because it is the distance from the surface where the potential is half the potential at the center.
137 Medium
137 of 372
Understanding +4 | -1
The relation between the intensity of the electric field of an electric dipole at a distance r from its centre on its axis and the distance r is where (r≫2l)r \gg 2l)
View answer and explanation
Correct option: D
E∝1r3E \propto \frac{1}{{{r^3}}}
Option D (E∝1r3E \propto \frac{1}{{{r^3}}}) is correct because the electric field due to a dipole at a distance r on its axis is inversely proportional to r cubed when r is much greater than 2l.
138 Medium
138 of 372
Application +4 | -1
If 7 charge qq are placed at corners of cube of edge length 3 m\sqrt{3} \mathrm{~m} then find potential at centre of cube:
View answer and explanation
Correct option: B
7q6πε0\frac{7 q}{6 \pi \varepsilon_0}
Option B (7q6πε0\frac{7 q}{6 \pi \varepsilon_0}) is correct because it is the electric potential at the center of the cube due to the 7 charges placed at its corners, calculated using the formula for electric potential due to a point charge.
139 Medium
139 of 372
Understanding +4 | -1
Which one statement is correct? A parallel plate air condenser is connected with a battery. Its charge, potential, electric field and energy are Q0,V0,E0Q_0, V_0, E_0 and U0U_0 respectively. In order to fill the complete space between the plates a dielectric slab is inserted, the battery is still connected. Now the corresponding values Q,V,EQ, V, E and UU are in relation with the initially stated as
View answer and explanation
Correct option: A
Q>Q0Q>Q_0
Option A (Q>Q0Q>Q_0) is correct because when a dielectric slab is inserted with the battery connected, the voltage remains constant, but the capacitance increases, leading to an increase in charge.
140 Medium
140 of 372
Application +4 | -1
The capacitance of a metallic sphere will be 1μF1\mu F, if its radius is nearly
View answer and explanation
Correct option: A
9km9km
Option A (9km9km) is correct because it is calculated using the formula for the capacitance of a sphere, solving for the radius.
141 Medium
141 of 372
Application +4 | -1
A series combination of N1\text{N}_1 capacitors (each of capacity C1\text{C}_1) is charged to potential difference 3 V.3 \mathrm{~V}. Another parallel combination of N2\mathrm{N}_2 capacitors (each of capacity C2\text{C}_2) is charged to potential difference V.\text{V}. The total energy stored in both the combinations is same. The value of C1\text{C}_1 in terms of C2\text{C}_2 is
View answer and explanation
Correct option: A
C2N1N29\frac{\mathrm{C}_2 \mathrm{N}_1 \mathrm{N}_2}{9}
Option A (C2N1N29\frac{\mathrm{C}_2 \mathrm{N}_1 \mathrm{N}_2}{9}) is correct because equating the energies stored in the series and parallel combinations allows us to solve for C1 in terms of C2, N1, and N2.
142 Medium
142 of 372
Understanding +4 | -1
The potential to which a conductor is raised, depends on
View answer and explanation
Correct option: C
Both (1) and (2)
Option C (Both (1) and (2)) is correct because the potential of a conductor depends on both the amount of charge and its geometry and size.
143 Medium
143 of 372
Application +4 | -1
A parallel plate condenser has a capacitance 50μF50\mu F in air and 110μF110\mu F when immersed in an oil. The dielectric constant ′k′'k' of the 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 capacitance in the medium to the capacitance in air.
144 Medium
144 of 372
Application +4 | -1
The equation of an equi-potential line in an electric field is y=2xy=2 x, then the electric field strength vector at (1,2)(1,2) may be:
View answer and explanation
Correct option: D
−8i^+4j^-8 \hat{i}+4 \hat{j}
Option D (-8i^\hat{i}+4j^\hat{j}) is correct because the electric field must be perpendicular to the equipotential line, and this option satisfies that condition.
145 Medium
145 of 372
Understanding +4 | -1
Assertion: For a point charge, concentric spheres centered at a location of the charge are equipotential surfaces.
Reason: An equipotential surface is a surface over which potential has constant value.
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 is correct because both the assertion and reason are true, but the reason doesn't explain why concentric spheres are equipotential surfaces for a point charge.
146 Medium
146 of 372
Understanding +4 | -1
When two conductors of charges and potentials C1,V1C_1, V_1 and C2,V2C_2, V_2 respectively are joined, the common potential will be:
View answer and explanation
Correct option: D
C1V1+C2V2C1+C2\frac{C_1 V_1+C_2 V_2}{C_1+C_2}
Option D (C1V1+C2V2C1+C2\frac{C_1 V_1+C_2 V_2}{C_1+C_2}) is correct because the common potential is found by equating the total charge before and after joining the conductors and solving for the final potential.
147 Medium
147 of 372
Analyzing +4 | -1
The electrostatic force between the metal plates of an isolated parallel plate capacitor CC having a charge QQ and area A{A}, is
View answer and explanation
Correct option: A
Independent of the distance between the plates
Option A is correct. The electrostatic force between the plates of an isolated parallel plate capacitor is independent of the distance between the plates because the charge Q is constant.
148 Medium
148 of 372
Understanding +4 | -1
Potential at any point inside a charged hollow
sphere:
View answer and explanation
Correct option: B
Is a constant
Option B (Is a constant) is correct because the electric field inside a charged hollow sphere is zero, implying that the potential is constant.
149 Medium
149 of 372
Application +4 | -1
12 J12 \mathrm{~J} of work has to be done against an existing electric field to take a charge of 0.01C0.01 \mathrm{C} from AA to BB. How much is the potential difference VB−VAV_B-V_A ?
View answer and explanation
Correct option: C
1200 V
Option C (1200 V) is correct because the potential difference is the work done per unit charge.
150 Medium
150 of 372
Understanding +4 | -1
The plates of a charged condenser is connected to a voltmeter. If the plates are moved apart, the reading of voltmeter will
View answer and explanation
Correct option: A
Increase
Option A (Increase) is correct because as the plates move apart, capacitance decreases, and since charge is constant, voltage increases.
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