Electric field of a charged rod

[twiztidmxcn]

hey

just looking for some help on an electric field question involving a rod of charge. here's the problem:

You have a charge, Q, uniformly distributed along a thin, flexible rod with length L. The rod is then bent into a semi-circle.

a) Find expression for electric field at center of semicircle
b) Evaluate field strength if L = 10cm, Q = 30nC.

The rod starts out straight and is then bent into a half circle.

We are also given the hint that: A small piece of arc length delta-s spans a small angle delta-theta = delta-s / R , where R is the radius.

Now, I realize that this problem has lots of symmetry, mostly where the x and y components of the electric field are concerned. I know that all the y components will cancel due to this symmetry and all that we're left with are the x components.

I'm attempting to use the equation of a rod of charge to derive something for the circle, but I am a bit stuck. Basically, I'm stuck at E = kq/r, r hat.

I believe that I can just use that equation, find r in terms of x and y (using triangles, pythagorean theorem) and then integrate in terms of x. I'm not quite sure about this though...

Any sort of help in the right direction would be much appreciated.

thanks
twiztidmxcn

 

[dextercioby]

Did you make an appropriate drawing...? Consider a small element of the chargel circular contour of length $dl$. What electric charge does it have...? Can you compute the electric field it creates using the formula of Coulomb and the particular (axially symmetric) geometry of the picture...?

Daniel.

 

[kyle8921]

Hello there,

It seems like you are on the right track with using the equation for the electric field of a rod of charge. In order to find the expression for the electric field at the center of the semicircle, you will need to integrate the equation over the entire length of the semicircle.

First, let's define a coordinate system where the center of the semicircle is at the origin and the x-axis is along the horizontal diameter. Let's also define a small piece of the semicircle with length delta-s and angle delta-theta, as given in the hint.

Now, using the equation for the electric field of a rod of charge, we can write the electric field at the center as:

E = ∫ k dq / r^2 r hat

Here, k is the Coulomb's constant, dq is the small amount of charge on the small piece of the semicircle, and r is the distance from the small piece to the center (which can be found using the pythagorean theorem).

Now, we can rewrite dq in terms of Q, the total charge on the rod, and delta-s, the length of the small piece:

dq = (Q/L) delta-s

Substituting this into the equation for electric field, we get:

E = ∫ (kQ/L) delta-s / r^2 r hat

Now, we need to find the expression for r in terms of x and y. Using the hint given, we can write:

r = R/2 + y

Where R is the radius of the semicircle and y is the vertical distance from the center to the small piece. Using the pythagorean theorem, we can also write:

r^2 = (R/2)^2 + (y-x)^2

Substituting this into the equation for electric field, we get:

E = ∫ (kQ/L) delta-s / [(R/2)^2 + (y-x)^2] r hat

Now, we can integrate over the entire semicircle by considering the limits of integration for y and delta-s. Since the semicircle is symmetric, we can integrate from 0 to R/2 for y and from 0 to pi for delta-s.

E = ∫(0 to pi) ∫(0 to R/2) (kQ/L) delta-s / [(R/2)^2 + (y-x)^