If the charge on a capacitor is increased by 2 colomb, the energy stored in it increases by $21 %$ . The original charge on the capacitor is

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Solution

The correct option is B 20
$\begin{matrix} A c c o r r d i n g t o q u e s t i o n . . . . . . . . . . . . . . . . . . . . . . . E n e r g y s t o r a g e i n a c a p a c i t o r : E = \frac{Q^{2}}{2 c} ∣ ∣ ∣ \begin{matrix} w h e r e, Q = c h a r g e o n t h e c a a c i t o r c = c a p a c i t a n c e o f t h e c a p a c i t o r \end{matrix} N o w, i f c a p a c i t a n c e o f t h e c a p a c i t o r i s c o n s t a n t . . . . . . t h e n, E = Q^{2} \Rightarrow \frac{E_{2}}{E_{1}} = {(\frac{Q_{2}}{Q_{1}})}^{2} \Rightarrow \frac{1.21 E_{1}}{E_{1}} = {(\frac{Q_{1} + 2}{Q_{1}})}^{2} \Rightarrow 1.1 = (\frac{Q_{1} + 2}{Q_{1}}) \Rightarrow Q_{1} + 2 = 1.1 Q_{1} \Rightarrow 0.1 Q_{1} = 2 ∴ Q_{1} = 20 C S o t h e c o r r e c t o p t i o n i s B . \end{matrix}$

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