First slide

Electrostatic potential energy

Question

Three particles each having a charge of 10 $μ C$ are placed at the corners of an equilateral triangle of side 10cm. The electrostatic potential energy of the system is (Given $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} N - m^{2} / C^{2}$

Easy

A

$Z e r o$
B

$I n f i n i t e$
C

$27 J$
D

$100 J$

Solution

For pair of charge $U = \frac{1}{4 π ε_{0}} . \frac{q_{1} q_{2}}{r}$
$U_{S y s t e m} = \frac{1}{4 π ε_{0}} [\frac{10 \times 10^{- 6} \times 10 \times 10^{- 6}}{10 / 100} + \frac{10 \times 10^{- 6} \times 10 \times 10^{- 6}}{10 / 100} + \frac{10 \times 10^{- 6} \times 10 \times 10^{- 6}}{10 / 100}]$
$= 3 \times 9 \times 10^{9} \times \frac{100 \times 10^{- 12} \times 100}{10} = 27 J$

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