Correct Answer - Option 2 : 8 J

__Concept__:

Energy stored in capacitor:

- A capacitor is a device 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 and

V = Electric potential difference

When capacitors are connected in** parallel,** the total capacitance is the sum of the individual capacitors' capacitances.

\({C_{eq}}(parallel) = {C_1} + {C_2} + {C_3} + \ldots + {C_4}\)

When capacitors are connected in **series,** the total capacitance is less than the least capacitance connected in series.

\(\frac{1}{{{C_{eq}(series)}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \ldots + \frac{1}{{{C_n}}}\)

**Calculations:**

C_{eq} = 4 μF, V = 400

When capacitances are connected in series

then, \(\frac{1}{{{C_{eq}}}} = \frac{5}{C}\)

C = 20 μF

When capacitances are connected in parallel

C_{eq}' = 20 × 5 = 100 μF

\(U = \frac{{100 \times {{10}^{ - 6}} \times 400 \times 400}}{2}\)

=** 8 J**