The gray charge distribution shown generates an electric field corresponding to
ID: 2308374 • Letter: T
Question
The gray charge distribution shown generates an electric field corresponding to the following equipotential surfaces. The potentials at points A and B are V_A = 3.0 V and V = 1.0 V. (a) On this diagram, sketch some of the electric field lines resulting from the charge distribution. Is the charge distribution positive or negative? (Yes. you have enough information to tell) (b) Where is the electric field strongest? Explain. (c) How much work would it take to move a Q - 0.50 C point charge along a straight line from B to A? (d) Now consider a semicircular path from B to A. To move the Q = 0.50 C charge along this path, would it take more work, less work, or the same work, as compared to part (c)? Explain. (e) Which takes more work: Moving charge Q from point C to point A, or moving it from point B to point A? Justify your answer.Explanation / Answer
(a) The field lines will point from A to towards B perpendicular to the equipotential surfaces as A is at higher potential than B .the charge distribution is negative as the potential is increasing as we move away from the charge distribution .
(b) The field will be strongest near point c where the equipotential surfaces are closest to each other
(c) The work done in moving charge 0.5.C from point B to point A is given by
dW = QdV = .50(3v - 1v) = 1N
(d) Since elecrostatic forces are conservative forces so the work done does not depend on path followed , hence thw workdone will be same as part c in this case also
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