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Potential energy of 2 charge $10 \mathrm{nC}$ each separated by a distance of $0.09 \mathrm{~m}$ in air is
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Verified Answer
The correct answer is:
$10 \mu \mathrm{J}$
Given $q_{1}=q_{2}=10 \mathrm{nC}=10 \times 10^{-9} \mathrm{C}$
$r=0.09 \mathrm{~m}$
Potential energy of two point charges is given by
$\begin{aligned}
U &=\frac{1}{4 \pi \varepsilon_{0}} \times \frac{q_{1} q_{2}}{r} \\
&=9 \times 10^{9} \times \frac{10 \times 10^{-9} \times 10 \times 10^{-9}}{0.09}=10^{-5} \mathrm{~J} \\
&=10 \mu \mathrm{J}
\end{aligned}$
$r=0.09 \mathrm{~m}$
Potential energy of two point charges is given by
$\begin{aligned}
U &=\frac{1}{4 \pi \varepsilon_{0}} \times \frac{q_{1} q_{2}}{r} \\
&=9 \times 10^{9} \times \frac{10 \times 10^{-9} \times 10 \times 10^{-9}}{0.09}=10^{-5} \mathrm{~J} \\
&=10 \mu \mathrm{J}
\end{aligned}$
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