Question

Two positive point charges of \(12 \mu \mathrm{c}\) and \(8 \mu \mathrm{c}\) are placed \(10 \mathrm{~cm}\) apart in air. The work done to bring them $4 \mathrm{~cm}$ closer is (A) Zero (B) \(4.8 \mathrm{~J}\) (C) \(3.5 \mathrm{~J}\) (D) \(-5.8 \mathrm{~J}\)

Short answer

The work done to bring the two point charges $4\mathrm{~cm}$ closer is approximately \(5.76 \mathrm{J}\). The given options contain an error, as none of them match the calculated result.

Step-by-step solution

Identify given values and the formula to use

The given values are: \(Q_1 = 12 \mu \mathrm{C} = 12 \times 10^{-6} \mathrm{C}\), \(Q_2 = 8 \mu \mathrm{C} = 8 \times 10^{-6} \mathrm{C}\), initial distance \(r_1 = 10 \mathrm{cm} = 0.1 \mathrm{m}\), and final distance \(r_2 = 6 \mathrm{cm} = 0.06 \mathrm{m}\). We will use the formula for the electrostatic potential energy: \[ U = \frac{kQ_1Q_2}{r} \]

Calculate the initial potential energy

Find the initial potential energy by plugging in the values into the formula: \[ U_1 = \frac{kQ_1Q_2}{r_1} \] \[ U_1 = \frac{(9 \times 10^9 \mathrm{Nm^2C^{-2}})(12 \times 10^{-6} \mathrm{C})(8 \times 10^{-6} \mathrm{C})}{0.1 \mathrm{m}} \] \[ U_1 = 8.64 \mathrm{J} \]

Calculate the final potential energy

Find the final potential energy by plugging in the values into the formula: \[ U_2 = \frac{kQ_1Q_2}{r_2} \] \[ U_2 = \frac{(9 \times 10^9 \mathrm{Nm^2C^{-2}})(12 \times 10^{-6} \mathrm{C})(8 \times 10^{-6} \mathrm{C})}{0.06 \mathrm{m}} \] \[ U_2 = 14.4 \mathrm{J} \]

Calculate the work done

Find the work done by calculating the difference between the final and initial potential energies: \[ W = U_2 - U_1 \] \[ W = 14.4 \mathrm{J} - 8.64 \mathrm{J} \] \[ W = 5.76 \mathrm{J} \] Since the work done is not equal to the given options, we can conclude that the exercise has an error in its options. The correct answer should be approximately \(5.76 \mathrm{J}\).