The capacitance of a variable capacitor connected to a 100 V battery is changed from \( 2\,\mu\mathrm{F} \) to \( 5\,\mu\mathrm{F} \). What is the change in the energy stored in it?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 1.5\times10^{-2}\,\mathrm{J} \)
The capacitor stays connected to the battery, so the potential difference V stays at 100 V while the capacitance changes. The energy stored is \( U=\frac{1}{2}CV^{2} \), so the change in energy is \( \frac{1}{2}(C_2-C_1)V^{2} \).
In SI units the data are:
\( C_1=2\,\mu\mathrm{F}=2\times10^{-6}\,\mathrm{F} \)
\( C_2=5\,\mu\mathrm{F}=5\times10^{-6}\,\mathrm{F} \)
\( V=100\,\mathrm{V} \)
The change in the energy stored is:
\( \begin{aligned} \Delta U &= \frac{1}{2}\,(C_2-C_1)\,V^{2} \\ &= \frac{1}{2}\times3\times10^{-6}\,\mathrm{F}\times(100\,\mathrm{V})^{2} \\ &= 1.5\times10^{-2}\,\mathrm{J} \end{aligned} \)
The energy increases, because the larger capacitance holds more charge at the same potential difference.