Two charges of 5nC and – 2nC, are placed at points (5 cm, 0, 0) and (23 cm, 0, 0) in a region of space. Where there is no other external field. Calculate the electrostatic potential energy of this charge system.

Solution

Given, $charge on particle 1 - q_{1} = 5 nC = 5 \times 10^{- 9} C c h a r g e o n p a r t i c l e 2 - q_{2} = - 2 nC = - 2 \times 10^{- 9} C$

The charges are placed on x-axis.

$distance between charges - r = 23 - 5 = 18 cm = 0.18 m$

$∴$ Electrostatic potential energy of the system of charge is,

$U = \frac{1}{4 {πε}_{0}} . \frac{q_{1} q_{2}}{r}$

$= \frac{9 \times 10^{9} \times 5 \times 10^{- 9} \times (- 2) \times 10^{- 9}}{0.18}$

$= \frac{- 9 \times 5 \times 2 \times 10^{- 9} \times 10^{2}}{18} U = - 5 \times 10^{- 7} J$

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Electrostatic Potential And Capacitance

Two charges of 5nC and – 2nC, are placed at points (5 cm, 0, 0) and (23 cm, 0, 0) in a region of space. Where there is no other external field. Calculate the electrostatic potential energy of this charge system.

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