How Do You Calculate the E-Field of a Semi-Circle with a Uniformly Charged Line?

[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\epsilonR₀

Any help with this would be greatly appreciated.

P.S. I'm new to latex.

 

[diazona]

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.

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.

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)

Actually dE, not E_i:
\mathrm{d}\mathbd{E} = \frac{1}{4\pi\epsilon_0} \frac{\mathrm{d}q}{R_0^2}\cos\theta
but anyway:

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\epsilonR₀

Any help with this would be greatly appreciated.

P.S. I'm new to latex.

I think your answer is off by a factor of \pi but the procedure is basically right.

 

[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\piR_o? 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\piR_o/R_o²)

And it cancels to what I said?

 

[rl.bhat]

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\piR_o? 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\piR_o/R_o²)

And it cancels to what I said?

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

 

[forty]

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

 

[Doc Al]

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

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

 

[forty]

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

dE = μ_ο/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?)

 

[Doc Al]

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

dE = μ_ο/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?)

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.)

 

[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!

 

[rl.bhat]

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

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.