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 \( U \) of a system of two point charges can be calculated using the formula: 

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

where:

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

Substitute all the given values into the formula:

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

Calculating step-by-step:

• The product of charges: \( 3 \times 10^{-6} \times -3 \times 10^{-6} = -9 \times 10^{-12} \)
• Substitute into the formula: \( U = \frac{9 \times 10^9 \times -9 \times 10^{-12}}{0.02} \)
• Calculate the numerator: \( 9 \times 10^9 \times -9 \times 10^{-12} = -81 \times 10^{-3} \)
• Divide by \( r \): \( \frac{-81 \times 10^{-3}}{0.02} = -4050 \times 10^{-3} \)
• Convert to joules: \( -4050 \times 10^{-3} \, \text{J} = -4.05 \, \text{J} \)

The electric potential energy of the system is \( -4.05 \, \text{J} \), which corresponds to the correct option.

Answer 2

Step 1: Use formula for electric potential energy of two point charges: 
\[ U = \frac{k q_1 q_2}{r} \] where \( q_1 = +3 \times 10^{-6} \, \text{C}, q_2 = -3 \times 10^{-6} \, \text{C}, r = 0.02 \, \text{m} \)

Step 2: Substitute the values. 
\[ U = \frac{9 \times 10^9 \cdot (3 \times 10^{-6}) \cdot (-3 \times 10^{-6})}{0.02} = \frac{-81 \times 10^{-3}}{0.02} = -4.05 \, \text{J} \]