Name: Electric Energy Density
Uploaded: 2019-07-19
Duration: 4 min 11 s
Description: There is an energy associated with the electrostatic field. The expression of this electric field energy density is derived using the energy stored in a para...

Question

Two closely spaced equipotential surfaces $A$ and $B$ with potentials $V$ and $V + δ V$ , (where $δ V$ is the change in $V$ ), are kept $δ l$ distance apart as shown in the figure. Deduce the relation between the electric field and the potential gradient between them. Write the two important conclusions concerning the relation between the electric field and electric potentials.

Accepted Answer

Work done in moving a unit positive charge through distance $δ l$ δl,

$E \times δ l = V_{A} - V_{B} = V - (V + δ V) = - δ V$

$E = \frac{- δ V}{δ l}$

The quantity $\frac{δ V}{δ l}$ is the rate of change of potential with distance and is called potential gradient. Thus, the electric field at any point is equal to the negative of the potential gradient at that point. The negative sign shows that the direction of the electric field is in the direction of decreasing potential. Moreover, the field is in the direction where this decrease is steepest.

(i) Electric field is in that direction in which the potential decrease is steepest.

(ii) The magnitude of electric field is equal to the change in the magnitude of potential per unit displacement (called potential gradient) normal to the equipotential surface at the given point.

Important Questions on Electrostatic Potential and Capacitance

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