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Derive a relation between electric field and potential |
Answer» Solution :CONSIDER two equipotential surfaces A and B with the potential difference dV between them as shown in figure. Let dl be the perpendicular DISTANCE between them and `vec(E )` be the electric field normal to these surfaces  The WORK done to MOVE a unit positive charge from B to A against the field `vec(E )` through a DISPLACEMENT `vec(dl)` is `dW= vec(E ).vec(dl)= E dl cos pi= - E dl` This is equal to the potential difference, therefore `dV= dW rArr dV= - E dl` `E= - (dV)/(dl)` |
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