Magnitude of the electric field

1. Jul 16, 2017

1. The problem statement, all variables and given/known data
http://imgur.com/a/g5acE

# 34

A plastic rod of length L = .24 cm is uniformly charged with a total charge of 12 x 10^-6 C.

The rod is formed into a semi circle with its center at the origin of the xy plane. What are the magnitude and direction of the electric field at the origin?

2. Relevant equations
kQ/r^2 = E

3. The attempt at a solution

Ok so can anyone tell me if this statement is correct:

1) Because I'm given L = .24 m, and the circumfrence of a circle is 2pir, since I have a semi circle I can set .24m = pi r, now r = .24m/pi

Going forward assuming that is true...:

The direction of the field is -j hat. Because the x components cancel out.

I consider a small piece of the rod I will call ds that has charge dQ. $dQ = \Lambda ds$ correct?

Anyways the magnitude of the electric field at a, E, is going to be 2dEsin(theta) because its getting its Y component from both sides of the Y axis.

so I have E = 2dEsin(theta)

dE = kdQ / r^2

so I have

E = 2*k*dq*sin(thetha) / r^2

but $dQ = \Lambda ds$

so

$E = (2*k*\Lambda * ds * sin(theta) ) / r^2$

now I integrate ds from 0 to (pi * r ) / 2

so I get (pi*r)/2 after integrating

$E = (*k*\Lambda * sin(theta) * pi * r ) / r^2$

pi * r * lambda are = Q, right?

so now I have

$E = kQ sin(theta) / r^2$

The reason sin(theta) wasn't in my integration was because the change of the angle has no impact on the E right?

But sin(theta) = y / r

so now I'm left with

E = kQy / r^3

But I have a feeling this is wrong. I can plug in for k, Q, and r^3, I'm not really sure what to do with y...

Last edited: Jul 16, 2017
2. Jul 16, 2017

I think I'm wrong when I say that sin(theta) doesn't have an impact on E, because it does? Since the electric field is stronger as y is closer to 0?

So at the step

$E = 2*k*\Lambda * sin(theta) / r^2$ , I find sin(theta) to be = y/r, so I get $E = 2*k*\Lambda * ds * y / r^3$

but how am I suppose to integrate y and ds ??

3. Jul 16, 2017

TSny

Yes.

Yes

Yes, assuming $\Lambda$ is the linear charge density.

Putting in the factor of 2 here is not a good idea. You are considering one element of charge dQ.

The left side would be $dE_y$ and you could add a negative sign to the right side to indicate that the y component is negative. Again, I would drop the factor of 2 here.

The arc length s and the angle θ are related. So, you should be able to write ds in terms of dθ and integrate with respect to θ. It's at this point that you can put in a factor of 2 and integrate over half the semicircle.

Last edited: Jul 16, 2017
4. Jul 17, 2017

ok so

$dE = dE_y sin(thetha)$

$dE_y = (k)(\Lambda) (ds ) / r^2$

$dE = (k)(\Lambda) (ds ) sin(theta) / r^2$

since s = (r)(theta) and ds = rdThetha

$E = (k)(\Lambda) (2)\int {sin(thetha) d(theta)} / r ^2$

int sin thetha = -cos evaluate from 0 to pi/2 you get one thus

$E = (k)(\Lambda)(2) / r^2$

correct??

set lambda = Q/L so i have

$E = (k)(\Lambda)(2) / (r^2) (L)$

I still feel like im doing something wrong :(, am I?

5. Jul 17, 2017

TSny

Shouldn't the symbols $dE_y$ and $dE$ be switched here?

Looks like you forgot the r in rdθ.

You are off by a factor of $r$ on the right side.
You can express L in terms of r and simplify. Should that be $\Lambda$ or $Q$ in your final result?
You almost have it.

Last edited: Jul 17, 2017