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Two point charges A and B of value of `+15 mu C and +9 mu C` are kept 18 cm apart in air. Calculate the work done when charge B is moved by 3cm towards A. |
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Answer» Correct Answer - `1.35 J` Here, `q_(1) = 15xx10^(-6)C, q_(2) = 9xx10^(-6)C` `r_(1) = 18xx10^(-2) m , r_(2) = (18 - 3) 10^(-2)m` work done = final P.E. - initial P.E. `w = (q_(1) q_(2))/(4pi in_(0) r_(2)) -(q_(1) q_(2))/(4pi in_(0) r_(1)) = (q_(1) q_(2))/(4pi in_(0)) [(1)/(r_(2)) - (1)/(r_(1))]` `= 9xx10^(9)xx15xx10^(-6)xx9xx10^(-6)xx9xx10^(-6) [(10^(2))/(15) - (10^(2))/(18)]` W = 1.35 joule |
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