A test charge `q_(0)` is moved without acceleration 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.

Explanation

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 angle to the electric field. So potential difference between P and Q is Ed.