Answer to Question #226089 in Electricity and Magnetism for danielel

Question

12 infinite lines of charge each with -λ are equally spaced and fixed around a circle of radius R. A proton (of charge +e) is located at the center of the circle. Now, the line charge at the 6 o'clock position is removed from the charge configuration leaving the others fixed. What is the magnitude of the net force on the proton?

Answer 1

If we remove the line charge at 6 o’clock position the force act on charge +e will be due to line charge at 12 o’clock. Because force due to line charges at position 1, 2, 3, 4, 5 o’clock position will be balanced by force due to line charges at 7, 8, 9, 10, 11 o’clock position respectively. Electric field at centre due to line charge at 12 o’clock, $\vec{E}$ then using Gauss’s Low

$\oint \vec{E} \times d \vec{S} = \frac{q}{ε_0} \\ = E \times 2 \pi Rl = \frac{-λl}{ε_0}$

$E = \frac{-λ}{2 \pi ε_0 R} \\ |E| = \frac{λ}{2 \pi ε_0 R} \\ Force = +e|E|=+e \frac{λ}{2 \pi ε_0 R}$