Consider two positive point charges *Q*_{1} and *Q*_{2}. Let’s have *Q*_{1} bound at position and *Q*_{2} initially at rest and infinitely far away. What does it take to move *Q*_{2} from to some position ?

Obviously, *Q*_{1} exerts a rightward electrical repulsion force (*F*_{e}) on *Q*_{2}. So a leftward external force (*F*_{ext}) must be applied on *Q*_{2} to bring it closer to *Q*_{1}. For some reason that I will explain later, let’s arrange for *F*_{ext} to match *F*_{e} in magnitude at every point between and . The total work done by *F*_{ext} during the journey is thus

Since *F*_{ext} matched *F*_{e} in magnitude throughout the journey, *Q*_{2} did not gain or lose any KE. So must represent the increase in EPE between and . But EPE is 0 at , so the EPE at is simply

To include the sign or not?

This formula is valid even for negative charges. So you should always include the signs of the charges when applying this formula. The formula will automatically calculate EPE to be positive when *Q*_{1} and *Q*_{2} are of the same signs, but negative when they are of opposite signs.

Example

*Q*_{A}, *Q*_{B} and *Q*_{C} are point charges of +1.1 pC, +2.2 pC and −3.3 pC, bound at the vertices of an equilateral triangle of side 3.0 um as shown in the diagram below. Calculate

a) total EPE of the system

b) the amount of work required to remove *Q*_{c} from the system (i.e. move it to a point at infinity).

Solution

a)

EPE is stored between 3 pairs of point charges.

Total EPE

b)

Before *Q*_{c} is removed: initial total EPE

After *Q*_{c} is removed: final total EPE

Work required

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**Explanation Video**