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.

Answer

`1.35 J`

Explanation

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