Question

The charge on a $48\mu F$ capacitor is increased from $0.1C$ to $0.5C$. The energy stored in the capacitor increases by

• A. $250J$
• B. $2500J$
• C. $2.5\times {10}^{6}J$
• D. $2.42\times {10}^{-6}J$

Answer 1

The correct option is B $2500J$

Given,

Initial charge on capacitor $Q_1=0.1C$

final charge on capacitor $Q_2=0.5C$

Capacitance $C=48\times {10}^{-6}F$

Increase in energy storage $=\frac{{Q_2}^2}{2C}-\frac{{Q_1}^2}{2C}=\frac{{0.5}^2}{2(48\times {10}^{-6})}-\frac{{0.1}^2}{2(48\times {10}^{-6})}=2500J$

Increase in energy storage is $2500J$