In the rectangle, shown below, the two corners have charges $q_{1} = -5 \mu C$ and $q_{2} = +2.0 \mu C$ . The work done in moving a charge $+3.0 \mu C$ from BR to $A$ is (take $1/4 \pi \varepsilon_0 = 10^{10} \mathrm{N} - m^2 / C^2$ )
Text Solution
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Work done $W = 3 \times 10^{-6} (V_{A} - V_{B});$ where
$V_A = 10^{10} \left[ \frac{-5 \times 10^{-6}}{15 \times 10^{-2}} + \frac{2 \times 10^{-6}}{5 \times 10^{-2}} \right] = \frac{1}{15} \times 10^{6} \text{ volt}$
and $V_B = 10^{10} \left[ \frac{\left( 2 \times 10^{-6} \right)}{15 \times 10^{-2}} - \frac{5 \times 10^{-6}}{5 \times 10^{-2}} \right] = - \frac{13}{15} \times 10^{6} \text{ volt}$
∴ ∴ $W = 3 \times 10^{-6} \left[ \frac{1}{15} \times 10^{6} - \left( \frac{13}{15} \times 10^{6} \right) \right]$ = 2.8 J
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