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

Show that the electric field at any point is equal to the negative of the potential gradient at that point.

Accepted Answer

Consider the electric field due to charge $+ q$ located at the origin $O$ . Let $A$ and $B$ be two adjacent points separated $b$ distance $d r$ .

The two points are so close that electric field $\vec{E}$ between them remains almost constant.

Let $V$ and $V + d V$ be the potentials at the two points.

The external force required to move the test charge $q_{0}$ (without acceleration) against the electric field $\vec{E}$ is given by

$\vec{F} = - q_{0} \vec{E}$

The work done to move the test charge from $A$ to $B$ is

$W = F . d r = - q_{0} E . d r$

Also, the work in moving the test charge from $A$ to $B$ is

$W = Charge \times potential difference$

Equating the two works done, we get $- q_{0} E . d r = q_{0} . d V$

or $E = - \frac{d V}{d r}$

The quantity $\frac{d V}{d r}$ 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.

Show that the electric field at any point is equal to the negative of the potential gradient at that point.

Important Questions on Electrostatic Potential and Capacitance

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