Question

Two charges of +3 μC and –3 μC are placed 2 cm apart in air. What is the electric potential energy of the system? (Take \( k = 9 \times 10^9 \) Nm\(^2\)/C\(^2\))

• A. \( -0.05 \, \text{J} \)
• B. \( -4.05 \, \text{J} \)
• C. \( +0.405 \, \text{J} \)
• D. \( -40.5 \, \text{J} \)

Hint:

Tip: Remember that potential energy between opposite charges is negative, indicating attraction.

Answer 1

The Correct Option is B

The electric potential energy \( U \) for two point charges is given by:

\( U = \frac{k \cdot q_1 \cdot q_2}{r} \)

Given values:

• \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \)
• \( q_1 = +3 \times 10^{-6} \, \text{C} \)
• \( q_2 = -3 \times 10^{-6} \, \text{C} \)
• \( r = 0.02 \, \text{m} \)

Substituting the values:

\( U = \frac{(9 \times 10^9) \cdot (3 \times 10^{-6}) \cdot (-3 \times 10^{-6})}{0.02} \)

Calculation steps:

• Product of charges: \( (3 \times 10^{-6}) \times (-3 \times 10^{-6}) = -9 \times 10^{-12} \)
• Substitute into formula: \( U = \frac{(9 \times 10^9) \times (-9 \times 10^{-12})}{0.02} \)
• Numerator calculation: \( (9 \times 10^9) \times (-9 \times 10^{-12}) = -81 \times 10^{-3} \)
• Division by \( r \): \( \frac{-81 \times 10^{-3}}{0.02} = -4.05 \)

The electric potential energy is \( -4.05 \, \text{J} \).