question_answer

Point charge \[q\] moves from point \[P\] to point \[S\] along the path \[PQRS\] (figure shown) in a uniform electric field \[E\] pointing coparallel to the positive direction of the \[X-\]axis. The coordinates of the points \[P,\,Q,\,R\] and \[S\] are \[(a,\,b,\,0),\ (2a,\,0,\,0),\ (a,\,-b,\,0)\] and \[(0,\,0,\,0)\] respectively. The work done by the field in the above process is given by the expression                                                     [IIT 1989]

A)                    \[qEa\]                            

B)            \[-qEa\]

C)                    \[qEa\sqrt{2}\]            

D)            \[qE\sqrt{[{{(2a)}^{2}}+{{b}^{2}}]}\]

Correct Answer: B

Solution :

           As electric field is a conservative field Hence the work done does not depend on path \ \[{{W}_{ABCD}}={{W}_{AOD}}\] \[={{W}_{AO}}+{{W}_{OD}}\] \[=Fb\text{cos 9}{{0}^{\text{o}}}+Fa\text{cos18}{{0}^{\text{o}}}=0+qEa\left( \text{1} \right)=qEa\]