Magnitude of the electric field

[Rijad Hadzic]

Homework Statement

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?

Homework Equations

kQ/r^2 = E

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. [itex] dQ = \Lambda ds [/itex] 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 [itex] dQ = \Lambda ds [/itex]

so

[itex] E = (2*k*\Lambda * ds * sin(theta) ) / r^2 [/itex]

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

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

[itex] E = (*k*\Lambda * sin(theta) * pi * r ) / r^2 [/itex]

pi * r * lambda are = Q, right?

so now I have

[itex] E = kQ sin(theta) / r^2 [/itex]

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^3But 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...

 

[Rijad Hadzic]

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

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

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

 

[TSny]

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

Yes.

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

Yes

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

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

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.

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

so

[itex] E = (2*k*\Lambda * ds * sin(theta) ) / r^2 [/itex]

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.

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

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.

 

[Rijad Hadzic]

Yes.

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.

ok so

[itex] dE = dE_y sin(thetha) [/itex]

[itex] dE_y = (k)(\Lambda) (ds ) / r^2 [/itex]

[itex] dE = (k)(\Lambda) (ds ) sin(theta) / r^2 [/itex]

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

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

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

[itex] E = (k)(\Lambda)(2) / r^2 [/itex]

correct??

set lambda = Q/L so i have

[itex] E = (k)(\Lambda)(2) / (r^2) (L) [/itex]

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

 

[TSny]

ok so

[itex] dE = dE_y sin(thetha) [/itex]

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

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

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

Looks like you forgot the r in rdθ.

int sin thetha = -cos evaluate from 0 to pi/2

you get one thus

[itex] E = (k)(\Lambda)(2) / r^2 [/itex]

correct??

You are off by a factor of ##r## on the right side.

set lambda = Q/L so i have

[itex] E = (k)(\Lambda)(2) / (r^2) (L) [/itex]

You can express L in terms of r and simplify. Should that be ##\Lambda## or ##Q## in your final result?
You almost have it.