Capacitors A and B are identical. Capacitor A is charged so it stores 4 J of energy and capacitor B is uncharged. The capacitors are then connected in parallel. The total stored energy in the capacitors is now:
A. 16 J
B. 8 J
C. 4 J
D. 2 J
E. 1 J

B. 8 J

Explanation:

From the question,

The voltage in each of the identical capacitor is given as

W = 1/2CV² Equation 1

Where Q = charge, V = potential difference, C = Capacitance of the capacitor.

make V the subject of the equation

V = √(2W/C) Equation 2

Given:  W = 4 J,

Substitute into equation 2 and find the value of V in terms of C

V = √(8/C)

Since both capacitor were connected in parallel,

The total potential difference = potential difference of on of the capacitor = √(8/C)

Total capacitance = 2C.

using equation 1,

W = 1/2(√(8/C))²×2C

W = 8 J.

Hence the right option is B. 8 J

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