Consider a test charge q0 as it moves from point A to point B in an electric field. The electric force on q0 at any point along the path is
\(\vec F\) = q0 \(\vec E\)
where \(\vec E\) is the field at that point.
The incremental work dW done by the field as q0 undergoes a displacement \(\vec {dl}\) along the path is
dW = \(\vec F\). \(\vec {dl}\)
= q0 \(\vec E\). \(\vec {dl}\)
= q0 E Dl
In the process, the charge q0 is moved from a higher potential to a lower potential, thereby losing potential energy. Therefore, the change in potential energy
dU = -dW = -q0 Edl
By definition, the change in electric potential is dv
If Ex, Ey and Ez are the rectangular components of \(\vec E\),
The quantity \(\cfrac{dV}{dl}\) is the rate at which the electric potential changes with distance and is called the electric potential gradient. The above equation thus shows that the magnitude of the electric field intensity at a point is equal to the negative of the potential gradient at that point.
[Note : (1) The negative sign shows that if we move in the direction of the electric field, the potential decreases. In the opposite direction, it increases. (2) If we draw equipotentials [see Unit 8.5] so that adjacent surfaces have equal potential differences, then in regions where the magnitude of \(\vec E\) is large, the equipotential surfaces are close together because the field does a relatively large amount of work on a test charge in a relatively small displacement. Conversely, in regions where the field s weaker, the equipotential surfaces are farther apart. (3) Potential gradient is an another name for electric field. (4) From the relation E = – dV/dx, we get another very common SI unit of electric field intensity, namely, the volt per metre (V/m).]
