Question

Two point charges of \( -5\,\mu C \) and \( 2\,\mu C \) are located in free space at \( (-4\,\text{cm}, 0) \) and \( (6\,\text{cm}, 0) \) respectively.
(a) Calculate the amount of work done to separate the two charges at infinite distance.
(b) If this system of charges was initially kept in an electric field \[ \vec{E} = \frac{A}{r^2}, \text{ where } A = 8 \times 10^4\, \text{N}\,\text{C}^{-1}\,\text{m}^2, \] calculate the electrostatic potential energy of the system.

Hint:

Work done to separate two charges equals the negative of potential energy. In a non-uniform field, calculate potential via integration before computing energy.

Answer 1

(a) Work Done to Separate Charges: Given: \[ q_1 = -5\,\mu C = -5 \times 10^{-6}\,\text{C}, \quad q_2 = 2\,\mu C = 2 \times 10^{-6}\,\text{C} \] \[ r = \text{distance between charges} = 6\,\text{cm} - (-4\,\text{cm}) = 10\,\text{cm} = 0.1\,\text{m} \] \[ W = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q_1 q_2}{r} \] \[ = 9 \times 10^9 \cdot \frac{(-5 \times 10^{-6})(2 \times 10^{-6})}{0.1} = -9 \times 10^9 \cdot \frac{10^{-11}}{0.1} = -0.9\,\text{J} \] Answer: \[ \boxed{W = -0.9\,\text{J}} \] (The work is negative because the charges attract, and external work is needed to separate them.) (b) Electrostatic Potential Energy in Field: Electric Field: \( \vec{E} = \frac{A}{r^2} \), with \( A = 8 \times 10^4 \,\text{N C}^{-1} \text{m}^2 \) Electric Potential at distance \( r \): \[ V = -\int \vec{E} \cdot dr = -\int \frac{A}{r^2} dr = \frac{A}{r} \] Calculate potential at positions of charges: \[ r_1 = 4\,\text{cm} = 0.04\,\text{m}, \quad r_2 = 6\,\text{cm} = 0.06\,\text{m} \] \[ V_1 = \frac{8 \times 10^4}{0.04} = 2 \times 10^6\,\text{V}, \quad V_2 = \frac{8 \times 10^4}{0.06} \approx 1.33 \times 10^6\,\text{V} \] Potential energy: \[ U = q_1 V_1 + q_2 V_2 = (-5 \times 10^{-6})(2 \times 10^6) + (2 \times 10^{-6})(1.33 \times 10^6) \] \[ U = -10 + 2.66 = \boxed{-7.34\,\text{J}} \]