Question

A capacitor is charged until its stored energy is 3 J and the charging battery is removed. Now another uncharged capacitor is connected across it and it is found that charge distributes equally. The final value of total energy stored in the electric fields is-

• A. 1.5 J
• B. 3 J
• C. 2.5 J
• D. 2 J

Answer 1

Answer: (A)

1.5 J

Answer 2

Energy stored in capacitor is, $\frac { 1 } { 2 }$CV² = 3 J

On connecting this capacitor to an uncharged capacitor since charge distributes equally, hence both capacitors are of same capacity Now common potential, V' = = Total energy stored in two capacitors is –

U' = $\frac { 1 } { 2 }$ CV'² + $\frac { 1 } { 2 }$ CV'²

= C

= $\frac { 1 } { 4 }$CV²

= $\frac { 3 } { 2 }$

= 1.5 J