Electrostatic Potential and Capacitance NEET Questions Chapter-wise

Q 1 :
A 12 pF capacitor is connected to a 50 V battery, the electrostatic energy stored in the capacitor in nJ is [2024]

15
7.5
0.3
150

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(1)

Capacitance, (C) $= 12$ $pF = 12 \times 10^{- 12}$ $F$

Voltage of Battery, (V) $= 50$ $V$

Energy stored in capacitor, (U) $= \frac{1}{2} C V^{2}$

$U = \frac{1}{2} \times (12 \times 10^{- 12}) {(50)}^{2} = 150 \times 10^{- 10}$ $J = 15 \times 10^{- 9}$ $J$ $= 15$ $nJ$

Q 2 :
If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then
A. the charge stored in it, increases.
B. the energy stored in it, decreases.
C. its capacitance increases.
D. the ratio of charge to its potential remains the same.
E. the product of charge and voltage increases.

Choose the most appropriate answer from the options given below. [2024]

A, B and E only
A, C and E only
B, D and E only
A, B and C only

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(2)

Since a battery is connected, so $V$ is constant.

$C = \frac{ε_{0} A}{d}; q = C V$ and, energy stored $U = \frac{1}{2} C V^{2}$

As $d$ decreases, $C$ increases, so $q$ increases. Energy stored in the capacitor also increases with increase in capacitance $C$ .
Hence, option (2) is correct.

Q 3 :
A capacitor of capacitance C = 900 $pF$ is charged fully by a 100 $V$ battery $B$ as shown in figure $(a)$ . Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance C = 900 $pF$ as shown in figure $(b)$ . The electrostatic energy stored by the system $(b)$ is [2022]

$4.5 \times 10^{- 6}$ J
$3.25 \times 10^{- 6}$ J
$2.25 \times 10^{- 6}$ J
$1.5 \times 10^{- 6}$ J

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(3)

Here, $C_{1} = C_{2} = 900$ $pF$

$V_{1} = 100$ $V, V_{2} = 0$ $V$

when $C_{1}$ and $C_{2}$ are joined, their common potential is given as,

$V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}} = \frac{900 \times 100 + 900 \times 0}{1800}$

$V = 50$ $V$

Electrostatic energy stored in system (b) is

$U = \frac{1}{2} (C_{1} + C_{2}) V^{2}$

$U = \frac{1}{2} \times 1800 \times 10^{- 12} \times 50 \times 50$

$U = 2.25 \times 10^{- 6}$ J

According to figure (b), charge on both the plate of capacitor is positive which is not possible.

Q 4 :
A parallel plate capacitor has a uniform electric field $' \vec{E}'$ in the space between the plates. If the distance between the plates is $' d'$ and the area of each plate $' A'$ , the energy stored in the capacitor is ( $ε_{0}$ = permittivity of free space) [2021]

$\frac{E^{2} A d}{ε_{0}}$
$\frac{1}{2} ε_{0} E^{2}$
$ε_{0} E A d$
$\frac{1}{2} ε_{0} E^{2} A d$

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(4)

The capacitance of a parallel plate capacitor of plate area $A$ and separation $d$ is given by

$C = \frac{ε_{0} A}{d}$

The potential is given by $V = E d$

Where $E$ is electric field.

Energy, $U = \frac{1}{2} C V^{2} = \frac{1}{2} \times \frac{ε_{0} A}{d} \times E^{2} d^{2}; U = \frac{1}{2} ε_{0} E^{2} A d$

Q 5 :
Two identical capacitors $C_{1}$ and $C_{2}$ of equal capacitance are connected as shown in the circuit. Terminals $a$ and $b$ of the key $k$ are connected to charge capacitor $C_{1}$ using a battery of emf $V$ volt. Now disconnecting $a$ and $b$ , the terminals $b$ and $c$ are connected. Due to this, what will be the percentage loss of energy? [2019]

75%
0%
50%
25%

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(3)

As we know that, loss of electrostatic energy,

$E_{loss} = \frac{1}{2} \frac{C_{1} C_{2}}{(C_{1} + C_{2})} V^{2} = \frac{1}{2} \times \frac{C^{2}}{2 C} V^{2}$

$= \frac{1}{2} (\frac{1}{2} C V^{2}) = \frac{1}{2} E$ $[∵ C_{1} = C_{2} = C]$

$∴$ $Percentage of loss of energy = \frac{\frac{1}{2} E}{E} \times 100 %$

$= \frac{1}{2} \times 100 % = 50 %$

Q 6 :
A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system [2017]

decreases by a factor of 2
remains the same
increases by a factor of 2
increases by a factor of 4

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(1)

When the capacitor is charged by a battery of potential $V$ , then energy stored in the capacitor,

$U_{i} = \frac{1}{2} C V^{2}$ ...(i)

When the battery is removed and another identical uncharged capacitor is connected in parallel

Common potential, $V' = \frac{C V}{C + C} = \frac{V}{2}$

Then the energy stored in the capacitor,

$U_{f} = \frac{1}{2} (2 C) {(\frac{V}{2})}^{2} = \frac{1}{4} C V^{2}$ ...(ii)

$∴$ $From eqns. (i) and (ii), U_{f} = \frac{U_{i}}{2}$

It means the total electrostatic energy of resulting system will decrease by a factor of 2.

Q 7 :
A parallel plate air capacitor of capacitance $C$ is connected to a cell of emf $V$ and then disconnected from it. A dielectric slab of dielectric constant $K$ , which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect? [2015]

The change in energy stored is $\frac{1}{2} C V^{2} (\frac{1}{K} - 1)$
The charge on the capacitor is not conserved.
The potential difference between the plates decreases $K$ times.
The energy stored in the capacitor decreases $K$ times.

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(2)

$q = C V \Rightarrow V = \frac{q}{C}$

Due to dielectric insertion, new capacitance

$C_{2} = C K$

Initial energy stored in capacitor, $U_{1} = \frac{q^{2}}{2 C}$

Final energy stored in capacitor, $U_{2} = \frac{q^{2}}{2 K C}$

Change in energy stored, $Δ U = U_{2} - U_{1}$

$Δ U = \frac{q^{2}}{2 C} (\frac{1}{K} - 1) = \frac{1}{2} C V^{2} (\frac{1}{K} - 1)$

New potential difference between plates $V' = \frac{q}{C K} = \frac{V}{K}$

Topic Question Set

Q 1 :
A 12 pF capacitor is connected to a 50 V battery, the electrostatic energy stored in the capacitor in nJ is [2024]

Q 4 :
A parallel plate capacitor has a uniform electric field $' \vec{E}'$ in the space between the plates. If the distance between the plates is $' d'$ and the area of each plate $' A'$ , the energy stored in the capacitor is ( $ε_{0}$ = permittivity of free space) [2021]

Q 6 :
A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system [2017]

Q 7 :
A parallel plate air capacitor of capacitance $C$ is connected to a cell of emf $V$ and then disconnected from it. A dielectric slab of dielectric constant $K$ , which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect? [2015]

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Topic Question Set

Q 1 : A 12 pF capacitor is connected to a 50 V battery, the electrostatic energy stored in the capacitor in nJ is [2024]

Q 4 : A parallel plate capacitor has a uniform electric field 'E→' in the space between the plates. If the distance between the plates is 'd' and the area of each plate 'A', the energy stored in the capacitor is (ε0 = permittivity of free space) [2021]

Q 6 : A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system [2017]

Q 7 : A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect? [2015]

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Q 1 :
A 12 pF capacitor is connected to a 50 V battery, the electrostatic energy stored in the capacitor in nJ is [2024]

Q 4 :
A parallel plate capacitor has a uniform electric field $' \vec{E}'$ in the space between the plates. If the distance between the plates is $' d'$ and the area of each plate $' A'$ , the energy stored in the capacitor is ( $ε_{0}$ = permittivity of free space) [2021]

Q 6 :
A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system [2017]

Q 7 :
A parallel plate air capacitor of capacitance $C$ is connected to a cell of emf $V$ and then disconnected from it. A dielectric slab of dielectric constant $K$ , which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect? [2015]