What will be the electric potential energy of a 7nC charge that is 2cm from a 20 nC charge?
Correct answer: C. 6.3 x 10^-5V
- A. 6.3 x 10^-1J
- B. 6.3 x 10^-15J
- C. 6.3 x 10^-5V
- D. 1.3 x 10^-5J
Explanation
The solution is as follows:The electric potential energy between two point charges is equal to work done in carrying charge to that point can be calculated using the formula:U = W = Q1V = Q2(KQ2/r) = KQ1Q2/rU = k * Q1 * Q2 / rwhere U is the electric potential energy, k is Coulomb's constant (9 × 10^9 N·m²/C²), Q1 and Q2 are the magnitudes of the charges, and r is the distance between the charges.In this case, Q1 = 7 nC = 7 × 10^-9 C, Q2 = 20 nC = 20 × 10^-9 C, and r = 2 cm = 0.02 m.Plugging these values into the formula, we get:U = (9 × 10^9 N·m²/C²) * (7 × 10^-9 C) * (20 × 10^-9 C) / (0.02 m)U = 6300 × 10^-18 JU = 6.3 × 10^ - 5 JTherefore, the electric potential energy of the 7 nC charge which is 2 cm from a 20 nC charge is 6.3 × 10^ - 5 J.
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