Questions: When an object with an electric charge of -40 μC is 5.0 cm away from an object with an electric charge of -30 μC, the electric potential energy of this system when the objects are 2.5 cm apart. Calculate the electric potential energy of this system when the objects are 2.5 cm apart. Be sure your answer has a unit symbol and the correct number of significant digits.

Solution Steps

We are given:

• Charge 1, \( q_1 = -40 \, \mu\text{C} = -40 \times 10^{-6} \, \text{C} \)
• Charge 2, \( q_2 = -30 \, \mu\text{C} = -30 \times 10^{-6} \, \text{C} \)
• Distance between the charges, \( r = 2.5 \, \text{cm} = 0.025 \, \text{m} \)

The electric potential energy \( U \) between two point charges is given by: \[ U = \frac{k \cdot q_1 \cdot q_2}{r} \] where \( k \) is Coulomb's constant, \( k = 8.988 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \).

Substitute the given values into the formula: \[ U = \frac{(8.988 \times 10^9) \cdot (-40 \times 10^{-6}) \cdot (-30 \times 10^{-6})}{0.025} \]

First, calculate the numerator: \[ (8.988 \times 10^9) \cdot (40 \times 10^{-6}) \cdot (30 \times 10^{-6}) = 8.988 \times 10^9 \times 1.2 \times 10^{-9} = 10.7856 \]

Then, divide by the distance: \[ U = \frac{10.7856}{0.025} = 431.424 \]

Final Answer