Question

Two-point charges, +3.0 μC, and +8.0 μC are separated by 0.30 m. The potential energy of this two charge system, relative to potential energy at infinite separation, is:  
1. 0.24 J
2. 0.36 J
3. 0.48 J
4. 0.72 J

Answer 1

Correct Answer - Option 4 : 0.72 J

Potential Energy:

• The amount of work done to move a charged particle from infinity to a point in an electric field is the energy needed to do so is known as potential energy for the system.
• This can easily be understood by the concept that as the charges of the two particles are the same there will be repulsion between them.
• Hence some work will be done to move the electron against the repulsive force.
• This work is converted to the potential energy of the system.
• The SI unit of electric potential energy is (J).

 

The potential energy (U) due to a charged particle at a distance r is given by:

\(U = \frac{1}{{4\pi {\epsilon_0}}} \times \frac{{{q_1}{q_2}}}{R}\) 

R = distance between charges or charged surface

E = Electric field due to charges

U = potential energy.

1/(4πϵ0) = constant = 9 × 109 N m2 /C

Calculation:

With q₁ = +3.0 μC, and q2 = +8.0 μC, the potential energy will be:

\(U = \frac{1}{{4\pi {\epsilon_0}}} \times \frac{{{3\times 10^{-6}}\times {8\times 10^{-6}}}}{0.3}\)

\(U = 9\times 10^{9} \times \frac{{{3\times 10^{-6}}\times {8\times 10^{-6}}}}{0.3}\)

U = 0.72 Joules