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A test charge q_(0) is movedwithoutacceleration from P to Q in a uniform electric field over the path shown in figure. The points P and Q are seperated by a distance d. Find the potential difference between P and Q. |
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Answer» <P> Solution :The test charge `q_(0)` FOLLOWING the path PRQ. For path PR.`V_(R ) -V_(P)= -int_(P)^(R ) barE.dbarl` `=-int_(P)^(R ) E.dlcost135^(0) = int_(P)^(R ) E.dl cos45^(0)` `=E cos45^(0) int_(P)^(R ) dl=E(PR) cos45^(0)` `=E(PQ)=Ed ( :. PQ=PR cos45^(0))` For path RQ `V_(Q)-V_(R ) = - int_(R )^(Q) barE. d barl= - int_(R )^(Q) E.dl cos90^(0) =0` `V_(Q)=V_(R )` No work is done in moving the charge at right angleto the electric field. So potential DIFFERENCE between P and Q is Ed. |
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