A \( 1\,\mu\mathrm{F} \) capacitor in a TV is charged to a potential difference of 4000 V. What is the energy stored in the capacitor?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 8 J
The work done in charging a capacitor is stored as electrostatic potential energy, \( U=\frac{1}{2}CV^{2} \) for capacitance C charged to a potential difference V.
\( C=1\,\mu\mathrm{F}=1\times10^{-6}\,\mathrm{F} \)
\( V=4000\,\mathrm{V} \)
\( \begin{aligned} U &= \frac{1}{2}CV^{2} \\ &= \frac{1}{2}\times1\times10^{-6}\,\mathrm{F}\times(4000\,\mathrm{V})^{2} \\ &= \frac{1}{2}\times10^{-6}\times1.6\times10^{7}\,\mathrm{J} \\ &= 8\,\mathrm{J} \end{aligned} \)