When charge `10 mu C` is shifted from inifinity to a point in an electric field, it is found that work done by eletrotatic forces is `10 muJ`. If the charge is doubled and taken again from inifinity to the same point without accelerating it, then find the amount of work done by electric field and against electric field.

Explanation

`(W_("ext"))_(oo P) = - (W_(e l))_(oo P) = (W_(e l)) _(p oo) = 10 mu J`
Because `Delta KE = 0`
`V_P = ((W_("ext"))_(oo p))/q = (10 mu J)/(10 mu C)= 1 V`
So if now the charge is doubled and taken from infinity, then
`1 = ((W_("ext"))_( oo P))/(20 mu C)` or `(W_(ext))_( oo P) = 20 mu J`
`(W_(e l))_( ooP) = -20 mu J`.