Correct Answer - Option 2 : 12 J
Concept:
A capacitor is a device used to store energy.
The process of charging up a capacitor involves the transferring of electric charges from one plate to another.
The work done in charging the capacitor is stored as its electrical potential energy.
The energy stored in the capacitor is
\(U = \;\frac{1}{2}\frac{{{Q^2}}}{C} = \frac{1}{2}C{V^2} = \frac{1}{2}QV\)
Where,
Q = charge stored on the capacitor
U = energy stored in the capacitor
C = capacitance of the capacitor
V = Electric potential difference
Calculation:
Capacitance (C) = 100 μF = 100 × 10-6 F
Applied voltage V1= 100 V and V2 = 500 V
The energy stored in the capacitor is
= \(\frac{1}{2} \times 100\times 10^{-6} \times \left( {V_{2\;}^2 - V_1^2} \right)\)
= \(\frac{1}{2} \times 100\times 10^{-6} \times \left( {{{500}^2} - {{100}^2}} \right)\)
= 12 J