Let’s consider a charge situated in an electric field and experiencing an electric force *F*_{e}. Suppose we apply an external force *F*_{ext} that is in opposite direction to *F*_{e}. By moving the charge some distance Δ*x* in opposite direction to *F*_{e}, we can move the charge to a position of higher EPE.

And if we purposely choose *F*_{ext} to be the same magnitude as *F*_{e}, we can keep the charge’s KE constant during the move. This will allow us to equate the (positive) work done by the external force to the increase in EPE, Δ*U*. In other words,

Rearranging, we get

If Δ*x *is small enough, then

So *F*_{ext} is equal to the potential energy gradient (at position A). But . So

This means that the electric force (at a point) is equal to the negative of the electric potential energy gradient (at that point)!

If we substitute and , we get

This means that electric field strength (at a point) is equal to the negative of the electric potential gradient (at that point).

Example

Two positive point charges of equal magnitude, *+Q* and *+Q*, produce electric fields and electric potentials around them.

Sketch the graphs showing the variation of the resultant *E* and total *V* at different position *x* along the line joining the two point charges.

Use the rightward is positive sign convention.

Solution

It is important to appreciate that these two graphs are related by .

Notice that the rate of change of *V* (with respect to distance) corresponds to the magnitude of *E* from: the steeper the potential gradient, the stronger the electric field. Also, notice that the null point in the *E-x* graph corresponds to the turning point in the *V-x* graph.

We can also tell the direction of *E* from the direction in which *V* is changing: *E* points in the direction in which *V* is decreasing.

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**Video Explanation**

Why Force is in the Direction of Decreasing EPE?

**Concept Test**