ELECTRIC POTENTIAL AND CAPACITANCE MCQs for NEET UG PHYSICS — Easy, Page 2

31 Easy

Understanding +4 | -1

A positively charged particle is released from rest in an uniform electric field. The electric potential energy of the charge

View answer and explanation

Correct option: C

Decreases because the charge moves along the electric field

Option C is correct because a positively charged particle moves along the electric field from a region of higher potential to lower potential, thus decreasing its electric potential energy.

Solution approach

Correct Answer + Core Reason: Decreases because the charge moves along the electric field. A positive charge moves from high to low potential, decreasing potential energy.

Step-by-Step Solution:

Step 1: Understand the relationship between electric field and electric potential.

Step 2: Electric field points in the direction of decreasing electric potential.

Step 3: A positive charge will move in the direction of the electric field, from a region of higher potential to a region of lower potential.

Step 4: Electric potential energy is given by $U = qV$ , where $q$ is the charge and $V$ is the electric potential.

Step 5: As the charge moves to a region of lower potential, its electric potential energy decreases.

Why other options are wrong:
- Option A: The potential energy changes as the charge moves in the electric field, even if the field is uniform.
- Option B: The charge moves along the electric field, but its potential energy decreases, not increases.
- Option D: The charge moves along the electric field, not opposite to it.

Key Concepts:
- Electric field
- Electric potential
- Electric potential energy

Memory Aids:
- Positive Charge: Moves to Lower Potential (Decreasing PE)

Quick Tips:
- Remember that positive charges move from high to low potential, and negative charges move from low to high potential.

32 Easy

32 of 45Question permalink

Analyzing +4 | -1

A parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved farther apart by pulling at them by means of insulating handles, then

View answer and explanation

Correct option: D

The voltage across the capacitor increases

Option D (The voltage across the capacitor increases) is correct because when the plates are moved apart after disconnecting the battery, the charge remains constant, and the capacitance decreases, leading to an increase in voltage.

Solution approach

Correct Answer + Core Reason: The voltage across the capacitor increases. Since Q is constant and C decreases, V must increase.

Step-by-Step Solution:

Step 1: Since the capacitor is disconnected from the battery, the charge $Q$ on the capacitor remains constant.

Step 2: The capacitance of a parallel plate capacitor is given by $C = \frac{{\varepsilon_0 A}}{d}$ , where $A$ is the area of the plates and $d$ is the distance between them. If $d$ increases, $C$ decreases.

Step 3: The voltage across the capacitor is given by $V = \frac{Q}{C}$ . Since $Q$ is constant and $C$ decreases, $V$ must increase.

Why other options are wrong:
- Option A: The energy stored in the capacitor increases, not decreases, since $U = \frac{1}{2} \frac{Q^2}{C}$ and C decreases.
- Option B: The capacitance decreases, not increases, as the distance between the plates increases.
- Option C: The charge on the capacitor remains constant, it does not decrease.

Key Concepts:
- Capacitance of a parallel plate capacitor: $C = \frac{{\varepsilon_0 A}}{d}$
- Relationship between charge, capacitance, and voltage: $Q = CV$
- Energy stored in a capacitor: $U = \frac{1}{2}CV^2 = \frac{Q^2}{2C}$

Memory Aids:
- Remember that when the battery is disconnected, the charge remains constant.

Quick Tips:
- Pay attention to whether the battery is connected or disconnected.
- Use the appropriate formula for energy based on whether charge or voltage is constant.

33 Easy

33 of 45Question permalink

Understanding +4 | -1

Two spheres of unequal radii are supplied with equal quantity of charges. The potential will

View answer and explanation

Correct option: B

Be more on the smaller sphere

Option B is correct because the electric potential is inversely proportional to the radius, so the smaller sphere will have a higher potential.

Solution approach

Correct Answer + Core Reason: Be more on the smaller sphere. Potential is inversely proportional to radius for a given charge.

Step-by-Step Solution:

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

Step 2: Since the spheres have equal charges, $Q_1 = Q_2 = Q$ .

Step 3: The potential of the first sphere is $V_1 = \frac{{Q}}{{4 \pi \epsilon_0 R_1}}$ .

Step 4: The potential of the second sphere is $V_2 = \frac{{Q}}{{4 \pi \epsilon_0 R_2}}$ .

Step 5: If $R_1 > R_2$ , then $V_2 > V_1$ .

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

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.

34 Easy

34 of 45Question permalink

Remembering +4 | -1

The electrostatic potential due to an electric dipole at a distance

r

varies as:

View answer and explanation

Correct option: B

\frac{1}{r^2}

Option B (

\frac{1}{r^2}

) is correct because the electrostatic potential due to an electric dipole varies inversely with the square of the distance.

Solution approach

Correct Answer + Core Reason: $\frac{1}{r^2}$ . The electrostatic potential due to an electric dipole varies inversely as the square of the distance.

Step-by-Step Solution:

Step 1: Recall the expression for the electrostatic potential due to an electric dipole.

Step 2: The electrostatic potential $V$ due to an electric dipole at a distance $r$ is given by: $V = \frac{p \cos{\theta}}{4 \pi \epsilon_0 r^2}$ , where $p$ is the dipole moment, $\theta$ is the angle between the dipole moment and the position vector, and $\epsilon_0$ is the permittivity of free space.

Step 3: From the expression, it is clear that the potential $V$ varies as $\frac{1}{r^2}$ .

Why other options are wrong:
- **Option A ( $\frac{1}{r^3}$ ):** This is incorrect. The potential due to a dipole varies as $\frac{1}{r^2}$ .
- **Option C ( $r$ ): This is incorrect. The potential does not increase with distance.
- Option D ( $\frac{1}{r}$ ): This is the potential due to a point charge, not a dipole.

Key Concepts:
- Electrostatic potential
- Electric dipole

Memory Aids:**
- Dipole potential: inversely proportional to $r^2$ .

Quick Tips:
- Remember the dependence of potential on distance for different charge distributions.

35 Easy

35 of 45Question permalink

Application +4 | -1

The minimum value of effective capacitance that can be obtained by combining 3 capacitors of capacitances

1\,\,pF,\,\,2\,\,pF

and

4\,\,pF

is

View answer and explanation

Correct option: A

\frac{4}{7}\,\,pF

Option A is correct. The minimum effective capacitance is obtained when the capacitors are connected in series.

Solution approach

Correct Answer + Core Reason: $\frac{4}{7}\,pF$ . Minimum capacitance is achieved in series.

Step-by-Step Solution:

Step 1: To obtain the minimum effective capacitance, connect the capacitors in series.

Step 2: Calculate the equivalent capacitance in series:
$\frac{1}{C_{eq}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{4} = \frac{4 + 2 + 1}{4} = \frac{7}{4}$
$C_{eq} = \frac{4}{7} pF$

Why other options are wrong:
- Option B: This is incorrect. This is the smallest individual capacitance, but the series combination will be even smaller.
- Option C: This is incorrect.
- Option D: This is the inverse of the correct answer.

Key Concepts:
- Capacitors in series
- Equivalent capacitance

Memory Aids:
- Series gives minimum capacitance.

Quick Tips:
- Remember that series connections result in lower overall capacitance.

36 Easy

36 of 45Question permalink

Understanding +4 | -1

If the plates of a charged capacitor are brought closer, then the

p

.(

d

) between the plates will

View answer and explanation

Correct option: A

Decrease.

Option A (Decrease) is correct because as the plates are brought closer, the distance decreases, and since the charge remains constant, the potential difference decreases.

Solution approach

Correct Answer + Core Reason: Decrease. When the plates are brought closer, the potential difference decreases.

Step-by-Step Solution:

Step 1: Recall the relationship between potential difference (V), charge (Q), and capacitance (C): $Q = CV$ .

Step 2: Rearrange the equation to solve for potential difference: $V = \frac{Q}{C}$ .

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

Step 4: Substitute the expression for C into the equation for V: $V = \frac{Q}{\frac{\epsilon_0 A}{d}} = \frac{Qd}{\epsilon_0 A}$ .

Step 5: Analyze the equation: If the charge Q remains constant and the distance d decreases, the potential difference V will decrease.

Why other options are wrong:
- Option B: Incorrect because the potential difference changes when the distance changes.
- Option C: Incorrect because the potential difference decreases when the distance decreases.
- Option D: Incorrect because the potential difference does not become zero.

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

Memory Aids:
- $Q = CV$
- $C = \frac{\epsilon_0 A}{d}$

Quick Tips:
- Remember the relationships between charge, capacitance, potential difference, and distance.

37 Easy

37 of 45Question permalink

Application +4 | -1

A

and

B

are two points in an electric field. If the work done in carrying 4 coulomb of electric charge from

A

to

B

is 16 Joule. Then the potential difference between

A

and

B

will be.

View answer and explanation

Correct option: C

4 Volt

Option C (4 Volt) is correct because the potential difference is the work done per unit charge, calculated as V = W/Q.

Solution approach

Correct Answer + Core Reason: 4 Volt. The potential difference is the work done per unit charge.

Step-by-Step Solution:

Step 1: Given: Work done $W = 16 J$ , Charge $Q = 4 C$ . Find: Potential difference V.

Step 2: Formula: Potential difference $V = \frac{W}{Q}$

Step 3: Substitute the given values: $V = \frac{16}{4}$

Step 4: Calculate: $V = 4 V$

Step 5: Final Answer: The potential difference between A and B is 4 Volt.

Why other options are wrong:
- Option A: Zero: This is incorrect because work is done to move the charge, implying a potential difference.
- Option B: 2 volt: This is incorrect. It might be a miscalculation.
- Option D: 16 Volt: This is incorrect. It might be a misunderstanding of the formula.

Key Concepts:
- Electric potential difference
- Work done in moving a charge

Memory Aids:
- V = W/Q

Quick Tips:
- Remember the definition of potential difference as work done per unit charge.

38 Easy

38 of 45Question permalink

Application +4 | -1

A condenser is charged to a p.d. of 240 volts, its energy is

2 \times {10^{ - 5}}\,\,{\text{joule}}{\text{.}}

If the battery is there and the space between the plates is filled up with a dielectric medium

\left( {{\varepsilon _r} = 10} \right),

its new energy is

View answer and explanation

Correct option: B

20 \times {10^{ - 5}}\,\,J

Option B (

20 \times {10^{ - 5}}\\ J

) is correct because the energy increases by a factor equal to the relative permittivity when the dielectric is inserted with the battery connected.

Solution approach

Correct Answer + Core Reason: $20 \times {10^{ - 5}}\,\,{\text{J}}$ . The energy increases by a factor of the dielectric constant when the battery remains connected.

Step-by-Step Solution:

Step 1: Recall the formula for energy stored in a capacitor.
$U = \frac{1}{2}CV^2$

Step 2: Calculate the initial capacitance.
$U_i = \frac{1}{2} C V^2 \implies 2 \times 10^{-5} = \frac{1}{2} C (240)^2 \implies C = \frac{4 \times 10^{-5}}{(240)^2} = \frac{4 \times 10^{-5}}{57600} \text{ F}$

Step 3: Calculate the new capacitance with the dielectric.
$C' = \epsilon_r C = 10 C = 10 \times \frac{4 \times 10^{-5}}{57600} \text{ F}$

Step 4: Calculate the new energy stored with the dielectric and the battery still connected (V remains constant).
$U' = \frac{1}{2} C' V^2 = \frac{1}{2} (10C) V^2 = 10 (\frac{1}{2} C V^2) = 10 U_i = 10 (2 \times 10^{-5}) = 20 \times 10^{-5} \text{ J}$

Why other options are wrong:
- Option A: $10 \times {10^{ - 5}}\,\,J$ is incorrect due to calculation errors.
- Option C: $40 \times {10^{ - 5}}\,\,J$ is incorrect due to calculation errors.
- Option D: $100 \times {10^{ - 5}}\,\,J$ is incorrect due to calculation errors.

Key Concepts:
- Energy stored in a capacitor
- Effect of dielectric on capacitance

Memory Aids:
- With battery connected, V is constant, so U increases by a factor of $\epsilon_r$ .

Quick Tips:
- If the battery is disconnected, the charge Q is constant, and U decreases by a factor of $\epsilon_r$ .

39 Easy

39 of 45Question permalink

Remembering +4 | -1

To store electric charge, we use

View answer and explanation

Correct option: D

A capacitor.

Option D (A capacitor) is correct because a capacitor is specifically designed to store electric charge.

Solution approach

Correct Answer + Core Reason: A capacitor. Capacitors are designed for storing electric charge.

Step-by-Step Solution:

Step 1: Consider the function of each component:
- A metal plate can hold some charge, but it's not designed for efficient charge storage.
- A magnetic dipole creates a magnetic field, not electric charge storage.
- An electric dipole consists of two equal and opposite charges, but it doesn't store a net charge.
- A capacitor is specifically designed to store electric charge.

Why other options are wrong:
- Option A: A metal plate can hold charge, but it's not the primary function.
- Option B: A magnetic dipole deals with magnetism, not charge storage.
- Option C: An electric dipole consists of separated charges, but does not store net charge.

Key Concepts:
- Capacitance
- Charge storage
- Electric and magnetic dipoles

Memory Aids:
- Capacitors store charge.

Quick Tips:
- Remember the function of basic electrical components.

40 Easy

40 of 45Question permalink

Application +4 | -1

A sphere P of radius 6 cm is charged to a potential of 12 V and is connected to an uncharged sphere Q of radius 3 cm. Determine the loss energy during the process.

View answer and explanation

Correct option: B

16 \times {10^{ - 11}}J

Option B (

16 \times {10^{ - 11}}J

) is correct because it is the calculated loss in energy after the two spheres are connected and reach the same potential.

Solution approach

Correct Answer + Core Reason: $16 \times {10^{ - 11}}J$ . The energy loss is calculated after charge redistribution between the spheres.

Step-by-Step Solution:

Step 1: Calculate the capacitance of each sphere.
$C_P = 4 \pi \epsilon_0 r_P = 4 \pi \epsilon_0 (0.06)$ and $C_Q = 4 \pi \epsilon_0 r_Q = 4 \pi \epsilon_0 (0.03)$

Step 2: Calculate the initial charge on sphere P.
$Q_P = C_P V_P = 4 \pi \epsilon_0 (0.06)(12) = 0.72 \pi \epsilon_0$

Step 3: Calculate the common potential after connection.
$V = \frac{Q_P}{C_P + C_Q} = \frac{0.72 \pi \epsilon_0}{4 \pi \epsilon_0 (0.06 + 0.03)} = \frac{0.72}{4(0.09)} = 2 \text{ V}$

Step 4: Calculate the final energy stored in the system.
$U_f = \frac{1}{2} (C_P + C_Q) V^2 = \frac{1}{2} [4 \pi \epsilon_0 (0.06 + 0.03)] (8)^2 = \frac{1}{2} [4 \pi \epsilon_0 (0.09)] (64) = 1.152 \pi \epsilon_0$

Step 5: Calculate the initial energy stored in sphere P.
$U_i = \frac{1}{2} C_P V_P^2 = \frac{1}{2} [4 \pi \epsilon_0 (0.06)] (12)^2 = 1.728 \pi \epsilon_0$

Step 6: Calculate the loss in energy.
$\Delta U = U_i - U_f = 1.728 \pi \epsilon_0 - 1.152 \pi \epsilon_0 = 0.576 \pi \epsilon_0 = 0.576 \pi (8.85 \times 10^{-12}) \approx 16 \times 10^{-11} \text{ J}$

Why other options are wrong:
- Option A: $12 \times {10^{ - 11}}J$ is incorrect due to calculation errors.
- Option C: $18 \times {10^{ - 11}}J$ is incorrect due to calculation errors.
- Option D: $24 \times {10^{ - 11}}J$ is incorrect due to calculation errors.

Key Concepts:
- Capacitance of a sphere
- Energy stored in a capacitor
- Charge redistribution

Memory Aids:
- Remember the formula for capacitance of a sphere: $C = 4 \pi \epsilon_0 r$

Quick Tips:
- When spheres are connected, they reach the same potential.

41 Easy

41 of 45Question permalink

Remembering +4 | -1

The energy stored in a condenser of capacity

C

which has been raised to a potential

V

is given by

View answer and explanation

Correct option: B

\frac{1}{2}C{V^2}

Option B (

\frac{1}{2}C{V^2}

) is correct because it is the standard formula for the energy stored in a capacitor.

Solution approach

Correct Answer + Core Reason: $\frac{1}{2}C{V^2}$ . This is the standard formula for energy stored in a capacitor.

Step-by-Step Solution:

Step 1: Recall the formula for the energy stored in a capacitor.\nThe energy stored in a capacitor is given by $U = \frac{1}{2}CV^2$ , where $C$ is the capacitance and $V$ is the potential difference.

Why other options are wrong:
- Option A: Incorrect. This is missing the factor of V.
- Option C: Incorrect. This is missing the factor of 1/2 and V squared.
- Option D: Incorrect. This is the inverse of the correct formula and is missing the square.

Key Concepts:
- Energy stored in a capacitor

Memory Aids:
- $U = \frac{1}{2}CV^2$

Quick Tips:
- Remember the standard formula for energy stored in a capacitor.

42 Easy

42 of 45Question permalink

Understanding +4 | -1

The ratio of electric fields due to an electric dipole on the axis and equatorial line at equal distance will be

View answer and explanation

Correct option: A

2

Option A (2) is correct because the electric field due to a dipole on the axis is twice that on the equatorial line at the same distance.

Solution approach

Correct Answer + Core Reason: 2. The electric field on the axial line is twice that on the equatorial line.

Step-by-Step Solution:

Step 1: Electric field on the axis of a dipole: $E_{axis} = \frac{{1}}{{4\pi \epsilon_0}} \frac{{2p}}{{r^3}}$

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

Step 3: Ratio: $\frac{{E_{axis}}}{{E_{equatorial}}} = \frac{{\frac{{1}}{{4\pi \epsilon_0}} \frac{{2p}}{{r^3}}}}{{\frac{{1}}{{4\pi \epsilon_0}} \frac{{p}}{{r^3}}}} = 2$

Why other options are wrong:
- Option B: 0: Incorrect. The ratio is 2.
- Option C: 0: Incorrect. The ratio is 2.
- Option D: 0: Incorrect. The ratio is 2.

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

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

Quick Tips:
- Remember the formulas for electric fields on the axial and equatorial lines.

43 Easy

43 of 45Question permalink

Application +4 | -1

An oil condenser has a capacity of

200\,\,\mu F.

The oil has a dielectric constant 5. When the oil leaks out its new capacity is

View answer and explanation

Correct option: A

40\,\,\mu F

Option A is correct. When the oil leaks out, the dielectric constant effectively becomes 1, so the new capacitance is the original capacitance divided by the dielectric constant.

Solution approach

Correct Answer + Core Reason: $40\\,\mu F$ . The capacitance decreases when the dielectric is removed.

Step-by-Step Solution:

Step 1: Recall the relationship between capacitance with and without a dielectric: $C = kC_0$ , where $C$ is the capacitance with the dielectric, $C_0$ is the capacitance without the dielectric, and $k$ is the dielectric constant.

Step 2: Given that the capacitance with the oil is $200 \mu F$ and the dielectric constant is 5, we have $200 \mu F = 5C_0$ .

Step 3: Solve for $C_0$ : $C_0 = \frac{200 \mu F}{5} = 40 \mu F$ .

Why other options are wrong:
- Option B: This is incorrect. It might be obtained by dividing the capacitance by 4 instead of 5.
- Option C: This is incorrect. This is the original capacitance with the oil.
- Option D: This is incorrect. This is the original capacitance multiplied by the dielectric constant squared.

Key Concepts:
- Dielectric constant and capacitance

Memory Aids:
- Dielectric increases capacitance by a factor of k.

Quick Tips:
- Remember that removing the dielectric decreases the capacitance.

44 Easy

44 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

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.

45 Easy

45 of 45Question permalink

Remembering +4 | -1

The electric potential at a point distant

\text{r}

from an electric dipole is proportional to

View answer and explanation

Correct option: B

\frac{1}{\mathrm{r}^2}

Option B (

\frac{1}{\mathrm{r}^2}

) is correct because the electric potential due to an electric dipole is inversely proportional to the square of the distance.

Solution approach

Correct Answer + Core Reason: $\frac{1}{\mathrm{r}^2}$ . The electric potential due to a dipole is inversely proportional to the square of the distance.

Step-by-Step Solution:

Step 1: Recall the formula for the electric potential due to an electric dipole at a distance $r$ from the dipole.

Step 2: The electric potential $V$ due to an electric dipole is given by: $V = \frac{kp\cos{\theta}}{r^2}$ , where $k$ is the electrostatic constant, $p$ is the dipole moment, $\theta$ is the angle between the dipole moment and the position vector, and $r$ is the distance from the dipole.

Step 3: From the formula, it is clear that the electric potential $V$ is proportional to $\frac{1}{r^2}$ .

Why other options are wrong:
- **Option A ( $\frac{1}{\mathrm{r}}$ ): This is the potential due to a point charge, not a dipole.
- Option C ( $\frac{1}{\text{r}^3}$ ):** The potential due to a quadrupole is proportional to $\frac{1}{r^3}$ .
- **Option D ( $\frac{1}{\mathrm{r}^4}$ ): This is not a standard dependence for electric potential.

Key Concepts:
- Electric dipole
- Electric potential

Memory Aids:**
- Dipole potential: inversely proportional to $r^2$ .

Quick Tips:
- Remember the formula for the electric potential due to a dipole.

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