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(a) Determine the electrostatic potential energy of a system consisting of two charges 7 µC and –2 µC (and with no external field) placed at (–9 cm, 0, 0) and (9 cm, 0, 0) respectively. (b) How much work is required to separate the two charges infinitely away from each other? (c) Suppose that the same system of charges is now placed in an external electric field E = A(1//r^(2)):A = 9 xx 10^(5) NC^(-1) . What would the electrostatic energy of the configuration be? |
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Answer» Solution :(a) `U = 1/(4pi epsilon_(0)) . (q_(1)q_(2))/( r) = 9 xx 10^(9) xx (7 xx -2) xx 10^(-12)/0.18 = 0.7 J` (B) `W = U_(2)-U_(1) =0-U = 0- (-0.7) = 0.7 J` ( c) The mutual interaction ENERGY of the two charges remains UNCHANGED. In addition, there is the energy of interaction of the two charges with the EXTERNAL electric field. We find, `q_(1) V(r_(1)) + q_(2)V(r_(2)) = A(7 muC)/(0.09 m) + A(-2 muC)/(0.09 m)` and the net electrostatic energy is `q_(1) V(r_(1)) + q_(2)V(r_(2)) + (q_(1)q_(2))/(4pi epsilonr_(12)) = A(7muC)/(0.09 m) + A(-2 muC)/(0.09 m) -0.7 J = 70-20-0.7 = 49.3 J` |
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