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Question
A point charge of +10 µC is placed at a distance of 20 cm from another identical point charge of +10 µC. A point charge of -2 µC is moved from point a to b as shown in the figure. Calculate the change in potential energy of the system? Interpret your result.
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Solution
q1 = 10μC = 10 x 10-6 C
q2 = 2μC = -2 x 10-6 C
distance, r = 5cm = 5 x 10-2 m
Change in potential energy,
`Delta "U" = (9 xx 10^9 xx 10 xx 10^-6 xx (- 2 xx 10^-6))/(5 xx 10^-2)`
= -36 × 1 × 109 × 10-12 × 102 = -36 × 10-1
∆ U = -3.6 J
Negative sign implies that to move the charge -2pC no external work is required. The system spends its stored energy to move the charge from point a to point b.
∆ U = -3.6 J, negative sign implies that to move the charge -2μC no external work is required. System spends its stored energy to move the charge from point a to point b.
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