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Hence ohtaln the work done in bringingcharge of 2 xx 10^(-9) C . From infinity to the point P. Does the answer depend on the path along which the charge is brought ? |
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Answer» <P> Solution :Let V(P) = and `V(oo) =0``:. V(P)-V(oo) = DeltaV=V=4xx10^(4)V` Now the work done in bringing a charge of `2xx10^(9)` C from infinity to the point P, W= q `DeltaV=qV` `= 2xx10^(-9) xx4xx10^(4)` `:. W= 8 xx10^(-5)J` No work done will be path independent. Any ARBITRARY infinitesimal path can be resolved into two PERPENDICULAR displacements. (1) Along parallel to displacement `vecr` and (2) Along perpendicular to displacement `vecr`. But work done corresponding to perpendicular component will be zero. |
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