# The charges on E = q and on F $\frac{q}{4}$ respectively. It is moved from points A to B one after another under an applied external electric field

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The charges on E = q  and on F $\frac{q}{4}$ respectively. It is moved from points A to B one after another under an applied external electric field. Let W be the work done in moving the charges each time. Then which of the following is true regarding the electrostatic potential of both charges?
1. VE = VF
2. V> VE
3. V< VF
4. $V_{E} \leq V_{F}$

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Correct Answer - Option 2 : V> VE

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 (QE) = q, Charge of F (QF) = $\frac{q}{4}$

Let VE = The potential difference when the charge E moved from A to B, VF = 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 VF > VE. Hence option 2 is the answer