Question

If the charge on the capacitor is increased by 2 coulomb, the energy stored in it increases by $21%$. The original charge on the capacitor is:
A. $10C$
B. $20C$
C. $30C$
D. $40C$

Answer 1

Hint:To solve the given problem we must know the expression for the energy stored in a given capacitor. Find the new stored energy in the capacitor with the help of the given data and calculate the original charge on it.

Formula used:
$E=\dfrac{{{q}^{2}}}{2C}$

Complete step by step answer:
A capacitor is an electronic component that stores charge. We understand the amount of charge that a given capacitor can store by its capacitance (C). When we apply a potential difference across a capacitor, the capacitor gets charged. As a result, the capacitor stores some amount of electric energy within it. The energy stored in a given capacitor of capacitance (C) and charge (q) is given as,
$E=\dfrac{{{q}^{2}}}{2C}$ …. (i).

It is given that the charge on the capacitor is increased by 2 coulomb. Therefore, the new charge on the capacitor is $q'=q+2$. It is also given that when the charge is increased, the energy stored in it increases by 21%. This means that the new stored energy is, $E'=\dfrac{121}{100}E$ …. (ii).
But we know that ${{E}^{'}}=\dfrac{q{{'}^{2}}}{2C}$.
Substitute the value of E’ in (ii).
$\dfrac{q{{'}^{2}}}{2C}=\dfrac{121}{100}E$.
Now, substitute the values of E from (i) and q’.
$\dfrac{{{(q+2)}^{2}}}{2C}=\dfrac{121}{100}\left( \dfrac{{{q}^{2}}}{2C} \right)$
$\Rightarrow 100{{(q+2)}^{2}}=121{{q}^{2}}$
$\Rightarrow 100{{q}^{2}}+400q+400=121{{q}^{2}}$
$\Rightarrow 21{{q}^{2}}-400q-400=0$

Now, use the quadratic formula to solve for q.
$\Rightarrow q=\dfrac{-(-400)\pm \sqrt{{{(-400)}^{2}}-4(21)(-400)}}{2(21)}$
Take 400 as common multiple inside the square root.
$ q=\dfrac{400\pm \sqrt{400(400+4(21))}}{42}$
$\Rightarrow q=\dfrac{400\pm \sqrt{400(484)}}{42}\\
\Rightarrow q=\dfrac{400\pm (22\times 20)}{42}\\
\Rightarrow q=\dfrac{400\pm 440}{42}$.
This means that $q=\dfrac{400+440}{42}=20C$ or $q=\dfrac{400-440}{42}= -0.95C$. However, we only consider the positive charge on the capacitor. Hence, $q=-0.95C$ is discarded. Therefore, the original charge on the capacitor is $q=20C$.

Hence, the correct option is B.

Note:Work has to be done to transfer charges onto a conductor, against the force of repulsion from the already existing charges on it. This work done to charges from one plate to the other is stored as potential energy of the electric field of the conductor and known as the energy stored in the capacitor.It is given as $E=\dfrac{C{{V}^{2}}}{2}$.