Explanation:
Given 6. (a) Determine the electro static potential energy of a system consisting of two charges 7 µ c and -2 µ c (and with no external field) placed at (-9 cm, 0, 0) and (9 cm, 0, 0) respectively. (b) How much work is required to separate the two charges infinitely away from each other? (c) Suppose that the same system of charges is now placed in an external electric field E = A (1/r^2). A = 9 x 10^5 cm^-2. What would the electrostatic energy of the configuration be?
- So two charges are q1 and q2
- So q1 = 7 μ C = 7 x 10^-6 C
- So q2 = - 2 μ C = 2 x 10^-6 C
- Distance between charges d = 9 – (- 9) = 18 x 10^-2 m
- So we have Electrostatic potential energy of the system will be
- E = 1/4πεo x q1q2 / r
- = 9 x 10^9 x 7 x 10^-6 x – 2 x 10^-6 / 18 x 10^-2
- = - 7 x 10^-1
- = - 0.7 J
- Now work done to separate the two charges infinitely away from each other
- W = E 2 – E 1
- = 0 – E
- = 0 – (- 0.7)
- = 0.7 J
- Now r1 = r2 = 9 cm = 0.09 m
- According to question, the system is placed in an electric field, electrostatic energy is
- = q1Er1 + q2Er2 + 1/4πεo q1q2 / r
- Now E = A.1 / r^2
- = q1 A / r1^2 x r1 + q2 A / r2^2 x r2 + 1/4πεo q1q2 / r
- = A q1 / r1 + A q2/r2 + 1/4πεo q1q2 / r
- = A q1 / r1 + A q2/r2 + E
- = 9 x 10^5 x 7 x 10^-6 / 0.09 + 9 x 10^5 x – 2 x 10^-6 / 0.09 – 0.7
- = 9 x 10^5 x 10^-6 / 0.09 (7 – 2) – 0.7
- = 0.9 / 0.09 x 5 – 0.7
- = 50 – 0.7
- = 49.3 J
Reference link will be
brainly.in/question/20186861