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.