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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


  • 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


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

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