Charge Rod Question: Electric Field at Point

[daveo123]

I've been having trouble with this one for a while, and I know where I'm stuck. Here's the problem:

There's a uniformly charged rod of length "L" along the x axis. There is also a point on the x-axis that is a distance of "a" away from the end of the rod. So the whole thing looks kind of like this. ----- o
I need to find the integral that will give me the magnitude of the electric field at that point. I started by using the charge density to come up with an expression for the infinitly small point charges that make up the rod. I got Q/L dx. So I put that into my equation for the field, dE, and I got

dE =k(Q/L dx) / r squared

where k is the constant from coulomb's law and r is the distance between each dx and the point I'm looking at. If the rest of this equation is correct (and I'm not sure that it is) my problem comes from finding an expression for r. The question asks me to "show with integration" that the field at the point is given by kq/a(a+L). I'm not sure how to represent the distance from the point charges to the point in the field that I'm measuring.

 

[daveo123]

Never mind, I found the answer elsewhere on the forum

 

[kyle8921]

Firstly, great job on starting with the charge density and using Coulomb's law to derive an expression for the infinitesimal electric field at a point on the x-axis. Your approach is correct. Now, to find the expression for the distance r, you can use the Pythagorean theorem.

Let's say the distance between each infinitesimal charge element (dx) and the point on the x-axis is represented by x. Then, the distance r can be represented by the hypotenuse of a right triangle with sides x and a. Using the Pythagorean theorem, we get r = √(x^2 + a^2).

Now, to find the total electric field at the point, we need to integrate the infinitesimal electric field expression over the length of the rod. This can be represented as:

E = ∫dE = ∫k(Q/L dx) / (√(x^2 + a^2))^2

= k(Q/L) ∫dx / (x^2 + a^2)

= k(Q/L) [1/a tan^-1(x/a)] from 0 to L

= k(Q/L) [tan^-1((L+a)/a) - tan^-1(a/a)]

= k(Q/L) (π/2 - 0)

= k(Q/L) (π/2)

= kQ/2L

= kq/a(a+L)

Therefore, the electric field at the point on the x-axis is given by kq/a(a+L). I hope this helps you in solving your problem. Keep up the good work!