Question

MI.1. A thin circular plastic ring carries a net charge that is uniformly distributed throughout the ring with a linear density of λ = 3.4 × 10-6 C/m. This ring is positioned parallel to a neutrally- charged infinite conducting plane such that its distance from the plane equals the radius (a) of the ring Fig.1]. It can be shown that the magnitude of the electric field on the axis of the this ring is given by: 20 (a+r2)3/2 where x is the distance from the ring along the axis. The field points along the axis of the ring, towards the ring (if the charge is negative), or away from it (if the charge is positive). A charged point particle is placed at the center of the ring. This particle has the same total charge as the ring, but it has the opposite sign. Find the magnitude (in Newtons) and direction of the net force on this particle.

Answer 1

Given, the linear density, lambda = 3.4 times 10^-6 c/m An electric field on the axis of a ring is, E^rightarrow (x) = lambda a x/x epsilon_0 (a^2 + x^2)^3/2 where, x is the distance from the center of the ring along the axis. We know, lambda = Q/2 pi a doublerightarrow Q = 2 pi a lambda The charge of a particle is = -Q = -2 pi a lambda The net electric field at the center of a ring is = 0 so, the net force on the particle due to a ring = ON The electric field due to an infinitely charged plate is, E = sigma/2 epsilon_0 where, sigma is net surface charge density sigma = sigma/2 + (-sigma/2) doublerightarrow sigma = 0 therefore, E = sigma/2 epsilon_0 E = 0 The electric field is zero. So, the force is zero. Therefore, the net force due to the ring and infinitely neutrally charged plate is zero. Hence, there is no direction