Two points charges of 10C each are kept at a distance of 3m in vacuum. Calculate their electrostatic potential energy.

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Solution

$q_{1} = q_{2} = 10 C; r = 3 m$
The electrostatic potential energy $= U = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r}$
$= \frac{9 \times 10^{9} \times 10 \times 10}{3}$
$= 3 \times 10^{11} J$

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Two points charges of 10C each are kept at a distance of 3m in vacuum. Calculate their electrostatic potential energy.

q1=q2=10C;r=3mThe electrostatic potential energy =U=14πε0q1q2r=9×109×10×103=3×1011J

$q_{1} = q_{2} = 10 C; r = 3 m$
The electrostatic potential energy $= U = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r}$
$= \frac{9 \times 10^{9} \times 10 \times 10}{3}$
$= 3 \times 10^{11} J$