Correct Answer - Option 2 : V

_{F }> V

_{E}
__CONCEPT__:

- The amount of work done in moving a unit positive charge in an electric field from infinity to that point without accelerating the charge against the direction of the electric field is electrostatic potential.
- The electrostatic potential is given by

⇒ \(V = \frac{W}{q}\)

Where V = Electrostatic potential, W = Work, and q = Charge

__CALCULATION__:

Given - Charge of E (Q_{E}) = q, Charge of F (Q_{F}) = \( \frac{q}{4}\)

Let V_{E} = The potential difference when the charge E moved from A to B, V_{F} = The potential difference when the charge F moved from A to B

- The potential difference when the charge E is moved from A to B can be written as

⇒ \(V_{E} = \frac{W}{q}\) --------(1)

- The potential difference when the charge F is moved from A to B can be written as

⇒ \(V_{F} = \frac{W}{\frac{q}{4}} = \frac{4W}{q}\) ------(2)

Equation 2 can be rewritten using equation 1 as

⇒ \( V_{F} = 4 V_{E}\) [ \([\because \ \frac{W}{q}= V_{E}]\)]

- From the above equation, it is clear that V
_{F} > V_{E}. Hence **option 2 is the answer**