# E-field of a semi circle.

1. Aug 1, 2009

### forty

A line charge of uniform density $$\lambda$$ forms a semi-circle of radius R0. Determine the magnitude and direction of the electric field intensity at the center of the semi-circle.

I won't bother with uploading a picture I'm pretty sure you can picture this without.

My trouble with this is I'm unsure whether I need to do a double integral.

So here is my working.

E = (1/4$$\pi$$$$\epsilon$$) q/r2 r

As the problem is symmetric only the cos($$\theta$$) components add. So for a small piece of charge dq

Ei = (1/4$$\pi$$$$\epsilon$$) dq/R02 cos($$\theta$$)

Now for this I have to sum up all the dq's but they are also all have a different angle. So do I do an integral for theta from -$$\pi$$/2 to $$\pi$$/2 ?

Anyway the answer I get is $$\lambda$$/2$$\epsilon$$R0

Any help with this would be greatly appreciated.

P.S. I'm new to latex.

Last edited: Aug 1, 2009
2. Aug 2, 2009

### diazona

First of all: no. Because it's a line of charge, i.e. a one-dimensional shape, you only need to do a single integral. (Think about it: if there were a double integral, what two variables would you integrate over?) If it were a two-dimensional surface, you might need to do a double integral; if it were a 3D volume, you might need a triple integral, etc.

Actually dE, not Ei:
$$\mathrm{d}\mathbd{E} = \frac{1}{4\pi\epsilon_0} \frac{\mathrm{d}q}{R_0^2}\cos\theta$$
but anyway:
I think your answer is off by a factor of $\pi$ but the procedure is basically right.

3. Aug 2, 2009

### forty

When I say double integral I mean in respect to all the dq's and all the d$$\theta$$'s.

When you sum up all the dq's to get the total charge isn't it $$\lambda$$$$\pi$$Ro? And the integral of cos($$\theta$$) from -pi/2 to pi/2 = 2?

So when I put it all together I get

E = (2/4$$\pi$$$$\epsilon$$)($$\lambda$$$$\pi$$Ro/Ro2)

And it cancels to what I said?

4. Aug 2, 2009

### rl.bhat

For a small element dL, charge dq = λ*dL and dL = R*dθ.
So the electric field at the center dE = μο/4π*λR*cosθ*dθ/R^2
Or dE = μο/4π*λ*cosθ*dθ/R. Τake the integration from 0 to π.

5. Aug 2, 2009

### forty

isn't the integral of cos(θ)dθ from 0 to pi equal to 0?

6. Aug 2, 2009

### Staff: Mentor

There's no double integral. You need to express dq in terms of the variable θ: dq = λrdθ

7. Aug 2, 2009

### forty

Yeah I see that now... but rl.bhat said

dE = μο/4π*λ*cosθ*dθ/R. Τake the integration from 0 to π.

isnt that integral 0? (isn't the integral of cos(θ)dθ from 0 to pi equal to 0?)

8. Aug 2, 2009

### Staff: Mentor

The integral should be from -π/2 to +π/2. (To avoid confusion, next time describe the orientation of the semicircle and how you define θ.)

9. Aug 2, 2009

### forty

I figured that after the first post I should of put up a picture. Thanks for clearing that up Doc Al (like always :) ). All makes sense now!

Thanks all!

10. Aug 2, 2009

### rl.bhat

Sorry. When you start with one end of the semicircle, the component of the field which add up is dEsinθ, rather dE*cosθ. Ιn that case the limit is 0 to pi.