Question

A parallel plate capacitor, at a capacity of $100\mu F$, is charged by a battery at $50{\text{V}}$. The battery remains connected and if the plates of the capacitors are separated so that the distance between them is reduced to half of the original distance, the additional energy given by the battery to the capacitor in ${\text{J}}$ is:
(A) $125 \times {10^{ - 3}}$
(B) $12.5 \times {10^{ - 3}}$
(C) $1.25 \times {10^{ - 3}}$
(D) $0.125 \times {10^{ - 3}}$

Answer 1

To solve this question, we need to use the formula for finding the initial energy stored in the capacitor. Then using the formula for the capacitance in terms of its geometrical parameters, we can find out the value of the final capacitance. Finally, using the same formula of the energy of a capacitor, we can find out the final energy, from which the value of the initial energy has to be subtracted to get the final answer.

Formula used: The formulae used for solving this question are given by
1. $E = \dfrac{1}{2}C{V^2}$, here $E$ is the energy stored in a capacitor of capacitance $C$ across which a voltage of $V$ is applied.
2. $C = \dfrac{{{\varepsilon _0}A}}{d}$, here $C$ is the capacitance of a parallel plate capacitor having the area of the plates as $A$ and the separation between the plates as $d$.

Complete step by step solution:
We know that the energy stored within a capacitor is given by
$E = \dfrac{1}{2}C{V^2}$ ……………….(1)
According to the question, the capacity of the capacitor is $C = 100\mu F = 100 \times {10^{ - 6}}F$, and the voltage of the battery is $V = 50{\text{V}}$. Substituting these above we get
$E = \dfrac{1}{2} \times 100 \times {10^{ - 6}} \times {50^2}$
$\Rightarrow E = 0.125{\text{J}}$ ……………..(2)
Now, according to the question, the distance between the plates of the capacitor is reduced to half. We know that the value of the capacitance is dependent upon this distance by
$C = \dfrac{{{\varepsilon _0}A}}{d}$
As can be seen from the above relation, the capacitance is inversely proportional to the distance between the plates. So when the distance is reduced to half, the capacitance becomes twice.
From (1) the energy is proportional to the capacitance. So when the capacitance becomes twice, the energy also becomes twice. So the final energy becomes
$E' = 2E$
From (2)
$E' = 2 \times 0.125{\text{J}}$
$\Rightarrow E' = 0.25{\text{J}}$ ……………...(3)
So the additional energy given by the battery to the capacitor is
$\Delta E = E' - E$
From (2) and (3) we have
$\Delta E = 0.25 - 0.125$
$\Rightarrow \Delta E = 0.125{\text{J}}$
This can also be written as
$\Delta E = 125 \times {10^{ - 3}}{\text{J}}$
Thus, the additional energy given by the battery to the capacitor is equal to $125 \times {10^{ - 3}}{\text{J}}$.

Hence, the correct answer is option A.

Note: After getting the value of the additional energy as $0.125{\text{J}}$, we should not tick the option D as your correct answer. We should beware of the factor of ${10^{ - 3}}$ which is there in each of the options. So we have to write the value of the energy as a multiple of this factor.