- #1

Rijad Hadzic

- 321

- 20

## 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...

Last edited: