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 =π2 and 14πε0=9×109Nm2/C2 ,is

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 μC from B to C is

$W = 5 \times 10^{- 6} (V_{C} - V_{B}) where$

$V_{B} = 9 \times 10^{9} \times \frac{100 \times 10^{- 6}}{0.4} = \frac{9}{4} \times 10^{6} V and V_{C} = 9 \times 10^{9} \times \frac{100 \times 10^{- 6}}{0.5} = \frac{9}{5} \times 10^{6} V So W = 5 \times 10^{- 6} \times (\frac{9}{5} \times 10^{6} - \frac{9}{4} \times 10^{6}) = - \frac{9}{4} J$