Question

A capacitor of capacity $C$ is connected with a battery of potential in parallel. The distance between its plates is reduced to half at once, assuming that the charge remains the same. Then to charge the capacitance upto the potential again, the energy given by the battery will be

• A. $C V ^ { 2 } / 4$
• B. $C V ^ { 2 } / 2$
• C. $3 C V ^ { 2 } / 4$
• D. $C V ^ { 2 }$

Answer 1

Answer: (D)

$C V ^ { 2 }$

Answer 2

Extra charge Q = (2CV – CV) = CV flows through potential V of the battery. Thus W = QV = $C V ^ { 2 }$