In the rectangle, shown below, the two corners have charges q1=−5μC and q2=+2.0μC. The work done in moving a charge +3.0μC from B to A is (take 1/4πε0=1010N-m2/C2)

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 $V_{B} = 10^{10} [\frac{(2 \times 10^{- 6})}{15 \times 10^{- 2}} - \frac{5 \times 10^{- 6}}{5 \times 10^{- 2}}] = - \frac{13}{15} \times 10^{6} volt$

$∴ W = 3 \times 10^{- 6} [\frac{1}{15} \times 10^{6} - (- \frac{13}{15} \times 10^{6})] = 2 .8 J$

In the rectangle, shown below, the two corners have charges $q_{1} = - 5 μC$ and $q_{2} = + 2 .0 μC$ . The work done in moving a charge $+ 3 .0 μC$ from B to A is (take $1 / 4 {πε}_{0} = 10^{10} N - m^{2} / C^{2}$ )

Solution:

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