There is a fixed positive charge Q at O and A and B are points equidistant from O. A positive charge q is taken slowly by an external agent from A to B along the line AC and then along the line CB .

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The total work done on the charge is zero

The work done by the electrostatic force from C to B is positive

The work done by electrostatic force in taking the charge from A to B is dependent on the actual path followed

The work done by the electrostatic force from A to C is negative

answer is A, B, C.

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Detailed Solution

$V_{A} = V_{B};$ work =0
The correct option is C The work done by the electrostatic force from C to B is positive.

A) is correct because the potentials of A and B are same as the points are equidistant from O. Therefore, there is no change in potential energy of the system. Thus, work done on the charge =0
B) is correct because the charge is going from low potential to high potential.
C) is correct because the charge is going from high potential to low potential.
D) is incorrect because electrostatic force is a conservative force. Hence, work done is independent of path.

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