Q3.13 Balancing electrostatic forces (K)
Two fixed charges, of +4q and -q are located at positions x=0 and x=L, respectively. Where should you place a charge of +2q along the x-axis for it to be in equilibrium.
Describe what would happen if the charge of +2q is displaced slightly to the left.
Hint
Reveal

Hint
In equilibrium means no net force. Sum the contributions to the overall force from each of the two fixed charges.Solution
Reveal

Solution
Let’s call the displacement of the moveable charge x=a. We can then calculate the repulsive force on it (F-) due to the +4q force at the origin:

And likewise the attractive force due to the -q charge at x=L

Equating these gives:

Thus


And so a=2L
If the charge is moved to the left, then both F- and F+ increase, but the attractive force increases more (it is closer to the 2q charge) so the charge continues to move to the left.
The opposite is true if it is moved to the right - it continues to move to the right (as both F- and F+ decrease this time, but the attractive force decrease more, so the particle moves to the right under the action of an imbalanced force. This is an example of an unstable equilibrium, more of which at the end of the course in Q7.1