Asked by  | 25th Jan, 2010, 09:56: PM

Expert Answer:

(a) All points in x-y plane where the potential is zero lies on a circle.

Let q be the test charge.

At all the points of zero potential i.e. (x,y) this test charge will not experience any electrostatic force.

2Qq/((x+3a)2+y2) = Qq/((x-3a)2+y2)

2((x-3a)2+y2) = (x-3a)2+y2

As we can see by further simplification of the above expression, that the locus will have quadratic terms in x and y. Further it's a circle.




Answered by  | 26th Jan, 2010, 09:51: AM

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