First slide
Coulumbs law

Two point charges A and B, having charges +Q and -Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes


In case I :

F=-14πε0Q2r2                                                      .........(i)  In Case II : QA=Q-Q4,QB=-Q+Q4 F'=14πε0Q-Q4-Q+Q4r2 =14πε034Q-34Qr2=-14πε0916Q2r2                  .......(ii)  From equations (i) and (ii), F'=916F 

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