Question

Show that the electric field intensity at a point in an electric field is equal to the -ve of the potential gradient of the field at that point.
i.e. E = \(-\frac{dV}{dr}.\)

Answer 1

Let P and Q be two points situated at a very small distance dr from each other in the electric field \(\vec E\)  of source charge +q lying at O Fig. The points P and Q and so close that the electric field \(\vec E\)  between P and Q is almost uniform as shown in Fig.

Let potential difference between P and Q = dV

By definition,

Potential difference between P and Q = Work done in moving a unit test charge from P to Q.

If a small positive test charge q₀ is moved from point P to Q, then