First slide

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 $\frac{1}{4 {πε}_{0}} = 9 \times 10^{9} {Nm}^{2} / C^{2}, is$

Moderate

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

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