Question

the graph in the figure shows the variation of the electric potential V(x) (in arbitrary units) as a function of the position x. Tell me everything you can about this situation, including what happens

if a negatively charge particle is placed at x = 1.5.

V(x) 2 -1.5-1-0.5 0.5 1 1.5 2x

Answer 1

Force on a particle is the first derivative of potential

$F=\frac{dV}{dx}$

Hence the force is zero at the point where the potential is minimum. For the given graph, the potential is minimum at x = 0. Hence the force is zero at x = 0. Therefore for a positive charge, the force is zero at x = 0 and hence x = 0 is the stable equilibrium point.

However, a negative charge flows from lower potential to higher potential. Hence the negatively charged particle placed at x = 1.5 will eventually move towards x = + infinity.