Question

Now, assume that we have divided the rod into many slices, and consider the small slice of width Δ r located at as shown below: Ax Question 2: By assuming that the slice in the picture below is a particle, calculate the total force it exerts on the particle g. Your answer should be in terms of Q. and Δι! Hints: . To find the charge of the slice, recall the thickness of the slice is Δ . To find the distance r between the slice and the particle with charge q, note that the slice is located at r and the particle is located at -3 . In the case when Δr 1/2,Q-80, q = 4.z-1, the result should be 20k; usel this to check if your expression is correct! Write down an integral that gives the total force, F, that the rod exerts on the particle. Your answer should be in terms of Q, g! Pay attention to the limits of integration Evaluate the integral you wrote down in order to compute the total force the rod exerts on the particle.

Answer 1

Charge on 2 = Q charge on 1 = Q/2 charge on delta x = Q/2 delta x r = (3 - x) so F = k Q_1 Q_2/r^2 F = kq(Q/2 delta x)/(3 - x)^2 = kQ q delta x/2(3 - x)^2 If delta x = 1/2, Q = 80, q = 4, x = 1 F = k times 80 times 4/2(3 - 1)^2 times 1/2 = 20 k to integrate we take delta x = dx so F = k Q q dx/2(3 - x)^2 \r\n F_0 = integral^2_0 F F = integral^2_0 k Q q dx/2(3 - x)^2 F_0 = kQq/2 integral^2_0 dx/(3 - x)^2 F_0 = kQq/2[1/3 - x]^2_0 F_0 = kQq/2[1/1 - 1/3] F_0 = kQq/3