First slide

Electrostatic potential

Question

A charge 10 esu is placed at a distance of 2 cm from a charge 40 esu and 4 cm from another charge -20 esu. The potential energy of the charge 10 esu is ( in erg) :

Easy

A

87.5
B

112.5
C

150
D

Zero

Solution

Potential energy = $\sum_{} \frac{q_{1} q_{2}}{r} = \frac{(10) (40)}{2} + \frac{(10) (- 20)}{4} = \frac{(10) (60)}{4} = 150 erg$

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