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A $12 \mathrm{pF}$ capacitor is connected to a $50 \mathrm{~V}$ battery. How much electrostatic energy is stored in the capacitor?
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Verified Answer
Given, $\mathrm{C}=12 \mathrm{pF}=12 \times 10^{-12} \mathrm{~F}$,
$$
\mathrm{V}=50 \mathrm{~V}, \mathrm{E}=\text { ? }
$$
By formula, electrostatic energy $\mathrm{E}=\frac{1}{2} \mathrm{CV}^2$
$$
=\frac{1}{2} \times 12 \times 10^{-12} \times(50)^2=1.5 \times 10^{-8} \mathrm{~J} \text {. }
$$
$$
\mathrm{V}=50 \mathrm{~V}, \mathrm{E}=\text { ? }
$$
By formula, electrostatic energy $\mathrm{E}=\frac{1}{2} \mathrm{CV}^2$
$$
=\frac{1}{2} \times 12 \times 10^{-12} \times(50)^2=1.5 \times 10^{-8} \mathrm{~J} \text {. }
$$
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