First slide

Electrostatic potential energy

Question

Two point charges $100 μC$ and $5 μC$ are placed at points A and B respectively with AB = 40 cm. The work done by external force in displacing the charge $5 μC$ from B to C, where BC = 30 cm, angle $ABC = \frac{π}{2}$ and $\frac{1}{4 {πε}_{0}} = 9 \times 10^{9} {Nm}^{2} / C^{2}$

Easy

A

9 J
B

$\frac{81}{20} J$
C

$\frac{9}{25} J$
D

$- \frac{9}{4} J$

Solution

Work done in displacing charge of 5 $μ$ C from B to C is
$W = 5 \times 10^{- 6} (V_{C} - V_{B})$ where
$V_{B} = 9 \times 10^{9} \times \frac{100 \times 10^{- 6}}{0 .4} = \frac{9}{4} \times 10^{6} V$
and $V_{C} = 9 \times 10^{9} \times \frac{100 \times 10^{- 6}}{0 .5} = \frac{9}{5} \times 10^{6} V$
So $W = 5 \times 10^{- 6} \times (\frac{9}{5} \times 10^{6} - \frac{9}{4} \times 10^{6}) = - \frac{9}{4} J$

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