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Two point charges 1 μC and -1 μC are brought to a distance of 1 meter from infinity. What is the change in electric potential energy?
1. 3 J
2. 9 J
3. 3 mJ
4. 9 mJ

1 Answer

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Correct Answer - Option 4 : 9 mJ

CONCEPT:

  •  Force one a charge q in an electric field is defined as 

⇒ F = Eq

  • The work done by this force in moving the charge by a distance d along the direction of electrostatic force (along the electric field) is

⇒ W = F.d = Eq.d

  • The work done by external force in moving this charge by a distance d along the direction of the electrostatic field is

⇒ Wext = - W = - Eq.d

  • By definition, the change in potential energy in moving the charge in an electrostatic field is equal to the work done by external forces

\(\Rightarrow \bigtriangleup U = W_{ext} = -Eq.d\)

  • Electric potential is equal to the amount of work done per unit charge by an external force to move the charge q from infinity to a specific point in an electric field

\(\Rightarrow V=\frac{W_{ext}}{q}\)

  • Therefore the relation between electric potential and electric potential energy is given by

\(\Rightarrow \bigtriangleup U = W_{ext} = Vq\)

EXPLANATION:

Given: q = q1 = q2 = 1 μC = 10-6 C and r = 1 m

  • The change in electric potential energy is given by

\(\Rightarrow \bigtriangleup U = W_{ext} = Vq = \frac {q^2}{4 \pi \epsilon_0r}=\frac{9 \times 10^9 \times 10^{-12}}{1^2}=9 \times 10^{-3}J = 9 mJ\)

  • Therefore option 4 is correct.

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