# Electric Field of a thin rod

1. Apr 17, 2010

### frostking

1. The problem statement, all variables and given/known data
We have a thin rod of length 10 cm uniformly charged with a total charge of + 40 nC . A small glass bead charged to + 6 nC is located 4 cm from the center of the rod. What is the force on the bead?

2. Relevant equations
E = integral of K dq /r^2 and Q = lambda L and therefore dq = lambda dy since y will be variable of length
L = length of rod d = distance from rod Lambda = Q/L

3. The attempt at a solution
The y component of the elec field is 0 due to symmetry.

Esub x = KQ / L [integral from - L/2 to + L/2 of d / (y^2 + d^2)^(3/2) dy

I can do this problem except for evaluating this integral. My book shows that this integral

d divided by (y^2 + d^2)^(3/2) dy when evaluated is y divided by [ d (y^2 + d^2)^(1/2) ] Can someone go over the steps to this integration PLEASE. I can take it after that and get a force but I can not figure out how this integral is done. THanks Frostking

2. Apr 18, 2010

### GRB 080319B

You need to make a trig substitution to solve this integral. With $$tan(\theta) = y/d$$, $$y = dtan(\theta)$$ and $$dy = dsec^{2}(\theta)d(\theta)$$. By substitution, and then using a trig identity on the denominator of the integral ( $$tan^{2}(\theta) + 1 = sec^{2}(\theta)$$, which is why we chose $$tan(\theta)$$ for the substitution ), you should get $$sec^{2}(\theta)/sec^{3}(\theta)$$, which is $$cos(\theta)$$. The evaluated integral is therefore $$sin(\theta)$$, which is equal to $$y/\sqrt{y^{2} + d^{2}}$$.

Last edited: Apr 18, 2010