First slide

Potential energy

Question

In the rectangle shown below, the two comers have charges q₁ = -5 μC and q₂ = +2.0 μC. The work done in moving a charge +3.0 μC from B to A is $(take 1 / 4 {πε}_{0} = 10^{10} N - m^{2} / C^{2})$

Moderate

A

2.8 J
B

3.5 J
C

4.5 J
D

5.5 J

Solution

$Work done W = 3 \times 10^{- 6} (V_{A} - V_{B}); where V_{A} = 10^{10} [\frac{(- 5 \times 10^{- 6})}{15 \times 10^{- 2}} + \frac{2 \times 10^{- 6}}{5 \times 10^{- 2}}] = \frac{1}{15} \times 10^{6} volt and V_{B} = 10^{10} [\frac{(2 \times 10^{- 6})}{15 \times 10^{- 2}} \frac{5 \times 10^{- 6}}{5 \times 10^{- 2}}] = - \frac{13}{15} \times 10^{6} volt ∴ W = 3 \times 10^{- 6} [\frac{1}{15} \times 10^{6} - (- \frac{13}{15} \times 10^{6})] = 2.8 J$

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