Question

answer it with steps

Answer 1

Hey There!
Here, there is a point charge Q. 
Electric Potential due to a point charge is given by :

$V=\frac{kQ}{r}$  
Now, the electron is initially at a distance r from the point charge. Even at the final position, it is at a distance r. 
The electron is actually moving on an equipotential surface. So, work done to move against the Electric Field of Point Charge is Zero.

However, there is still the Gravitational Force acting. 
The whole setup is in a vertical plane. On moving from A to B, the electron covers a vertical displacement of 2r.

In other words, electron loses a potential energy of mgh = mg(2r) = 2mgr

So, Net Work Done = Change in PE = -2mgr

Here, negative sign indicates loss in energy.
So, finally, your answer is Option 1) - 2mgr



Hope it helps,
Purva
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