# 12.5.1 Energy and Potential Gradient

Let’s consider a charge situated in an electric field and experiencing an electric force Fe. Suppose we apply an external force Fext that is in opposite direction to Fe. By moving the charge some distance Δx in opposite direction to Fe, we can move the charge to a position of higher EPE.

And if we purposely choose Fext to be the same magnitude as Fe, 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, ΔU. In other words,

$\displaystyle \Delta U={{F}_{{ext}}}\Delta x$

Rearranging, we get

$\displaystyle {{F}_{{ext}}}=\frac{{\Delta U}}{{\Delta x}}$

If Δx is small enough, then

$\displaystyle \displaystyle {{F}_{{ext}}}=\frac{{dU}}{{dx}}$

So Fext is equal to the potential energy gradient (at position A). But $\displaystyle {{F}_{{ext}}}=-{{F}_{e}}$. So

$\displaystyle \displaystyle {{F}_{e}}=-\frac{{dU}}{{dx}}$

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 $\displaystyle {{F}_{e}}=qE$ and $\displaystyle U=qV$, we get

\displaystyle \begin{aligned}qE&=-\frac{{d(qV)}}{{dx}}\\E&=-\frac{{dV}}{{dx}}\end{aligned}

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 $\displaystyle E=-\frac{{dV}}{{dx}}$.

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.

Video Explanation

Why Force is in the Direction of Decreasing EPE?

Concept Test

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