Two charges `+Q` each are fixed at points C and D. Line AB is the bisector line of CD. A third charge `+q` is moves from A to B, then from B to C

Options

A. From A to B electrostatic potential energy wil decrese
B. from A to B electrostastic potential ennergy wil increase
C. From B to C electrostatic potential energy will increase
D. From B to C electrostatic potential energy will decrease

Answer

B, C

Explanation

Along the line `AB` charge `q` is at unstable equililbrium position at B (when displaced from B along `AB` net force on it is away from B, whereas force at B is zero). Hence potential energy at B is maximum.
Along `CD` equilibrium of q is stable. Hence, potential energy at B is minimum along `CD`.