Two charges $4 \mu \mathrm{C}$ and 7 mC are placed 4.5 cm apart. The mutual P.E. of the system is\

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Professor PC Thomas Classes · Answer

PE of two charges $=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r}$\ $=\frac{9 	imes 10^{9} 	imes\left(4 	imes 10^{-6}ight)\left(7 	imes 10^{-3}ight)}{4.5 	imes 10^{-2}}$\