Two charges –q each are separated by distance 2d. A third charge + q is kept at midpoint O. Find potential energy of + q as a function of small distance x from O due to – q charges. Sketch P.E. v/s x and convince yourself that the charge at O is in an unstable equilibrium.

Question

Two charges –q each are separated by distance 2d. A third charge + q is kept at midpoint O. Find potential energy of + q as a function of small distance x from O due to – q charges. Sketch P.E. v/s x and convince yourself that the charge at O is in an unstable equilibrium.

Philoid · Accepted Answer

Given

Distance between charges = 2d

Magnitude of charges at ends = -q

Magnitude of charge at O = +q

The potential energy between two charges is given by the equation

Since the third charge +q is displaced form the point O by x distance, the total potential energy of the system will be,

PE vs x

We know that that the net force on a charge and the potential energy are related by the equation,

For equilibrium, the net force on the charge must be zero, therefore

Now,

On solving for x we get x=0 or point O. Thus, the system will be at equilibrium when the charge +q will be at O. The system will be stable at x=0 if there is minimum potential energy at that point, and for minimum potential energy,

Now,

At x=0

Since, d is positive,

Therefore, the charge at O will be in unstable equilibrium. This can also be seen from the P.E vs X graph where the potential energy has maxima at the point O.