Question

Potential energy of two charges $10 \, nC$ each separated by a distance of 0.09 m in air is

• A. $ 10 \mu J $
• B. $ 1 \mu J $
• C. $ 10 m J $
• D. $ 10 J $

Answer 1

The Correct Option is A

Potential energy of two charge system is given by
$U= \frac{1}{4 \pi e_{0} e_{r}} \times\frac{q_{1}q_{2}}{r} $
Here, $ q_{1 } = q_{2} = 10 nC = 10\times 10^{-9}C $
$ \frac{1}{4 \pi e_{0} } = 9 \times 10^{9} N m^{2} C^{-2} r = r = 0.09 m $
$ \therefore \, U = 9 \times 10^{9} \times \frac{10 \times 10^{-9} \times 10 \times 10^{-9}}{0.09} = 10 \mu J$

Answer 2

The electric potential energy of a charge in the electric field due to any charge is given by the work done by an external force in bringing a test charge from infinity to that point in the electric field.

Potential Energy of a System of Two Charges

Consider two point charges q₁ and q₂ initially lying at infinity.

Work done in bringing a charge q from infinity to a point where electric potential due to any charge is V is given by

W = qV

Initially when charge q₁ is brought from infinity to a particular point, there is no work is done in bringing it because at that point no electric potential is presence due any other charge i.e. V = 0 at that point. Therefore

W₁ = q₁ V = q₁ x 0 = 0

Now when charge q₂ is brought from infinity to a point at distance r from the charge q₁, the electric potential is present at that point due to charge q₁ that is given by

V = 1/4πϵ₀ q₁/r

Therefore, work done in bringing charge q₂ is given by

W₂ = q₂ x V = q₂ x 1/4πϵ₀ q₁/r

⇒ W₂ = 1/4πϵ₀ q₁q₂/r

Net work done in bringing both charges from infinity is given by

W = W₁ + W₂

⇒ W = 0 + 1/4πϵ₀ q₁q₂/r

⇒ W = 1/4πϵ₀ q₁q₂/r

This work done is stored as electric potential energy (U) of the system of the two charges. Hence

U = 1/4πϵ₀ q₁q₂/r