Question

Two equal positive point charges of 1 μC charge are kept at a distance of 1 metre in air. The electric potential energy of the system will be.

• A. 1 joule
• B. 1 eV
• C. \( 9 \times 10^{-3} \) joule
• D. zero

Hint:

The electric potential energy between two charges is directly proportional to the product of their charges and inversely proportional to the distance between them.

Answer 1

The Correct Option is C

Step 1: Formula for Electric Potential Energy.
The formula for the electric potential energy of two point charges is given by: \[ U = \frac{{k \cdot q_1 \cdot q_2}}{{r}} \] where \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \) is Coulomb's constant, \( q_1 = q_2 = 1 \, \mu C = 1 \times 10^{-6} \, C \), and \( r = 1 \, \text{m} \).
Step 2: Substituting the values.
\[ U = \frac{{9 \times 10^9 \cdot (1 \times 10^{-6})^2}}{{1}} = \frac{{9 \times 10^9 \cdot 10^{-12}}}{{1}} = 9 \times 10^{-3} \, \text{joule} \] Conclusion: The electric potential energy of the system is \( 9 \times 10^{-3} \) joule.