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Two charges +Q each are fixed at points ...

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

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

Text Solution

Verified by Experts

The correct Answer is:
B, C
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`.
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