First slide

Potential energy

Question

A charge of 10 e.s.u. is placed at a distance of 2 cm from a charge of 40 e.s.u. and 4 cm from another charge of 20 e.s.u. The potential energy of the charge 10 e.s.u. is (in ergs)

Easy

A

87.5
B

112.5
C

150
D

250

Solution

$Energy = \frac{10 \times 40}{2} + \frac{10 \times 20}{4} = 250 erg$

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