Finding Electric Field of Charged Dielectric Sphere

[khfrekek1992]

Homework Statement

There is a charged dielectric sphere of radius R. I center it at the origin of an x,y,z cartesian system. It has a charge density rho=3Qcos(theta)/4*Pi*R^3. Q and R are given.

I need to find the Electric field at a point that is outside the sphere.

Homework Equations

int(E*dA)=Q_in/Epsilon0

The Attempt at a Solution

int(E*dA)=Q_in/Epsilon0
E(4*Pi*r^2)=(rho)(4/3)*Pi*r^3)/epsilon0
E=3Qcos(theta)r/12*Pi*epsilon0*R^3 in the r-hat direction

Is this right? Dont I have to figure out the field in the x-hat, y-hat, and z-hat directions? How does this work if my E is in the r-hat direction?

Thanks in advance!

 

[SammyS]

It appears that you are trying to use Gauss's Law to find the electric field. The charge density depends on θ, so there isn't enough symmetry for Gauss's Law to be of help here.

It looks like you will have to use direct integration to find either electric field, E, or electric potential, V.

 

[khfrekek1992]

Thanks for the reply. I tried integrating rho directly earlier, but always got 0 because sin evaluated from 0 to Pi is always 0. :(

 

[khfrekek1992]

Find the Electric field at a point outside a charged sphere

Homework Statement

There is a charged dielectric sphere of radius R. I center it at the origin of an x,y,z cartesian system. It has a charge density rho=3Qcos(theta)/4*Pi*R^3. Q and R are given.

I need to find the Electric field at a point that is outside the sphere.

Homework Equations

E=-(grad)V
V=kq/r

The Attempt at a Solution

I integrated rho over the volume of the sphere to get q=(Pi/2)Qcos(theta)
then used V=q/4*pi*epsilon0r to get v=(Qcos(theta))/(8epsilon0*r)

How do i split this voltage into x,y, and z components so i can take the gradient to get E at a certain (x,y,z) point?

Thanks in advance

 

[SammyS]

This looks like the same problem that you posted in https://www.physicsforums.com/showthread.php?t=535920".

 

[khfrekek1992]

Yeah haha it is the same but I am doing the problem completely different now(from your suggestion) and am now stuck at a different part :/

 

[SammyS]

Yes, the net charge is indeed zero. Integrating the cosine from 0 to π does give zero.

That doesn't mean that the electric potential and/or the electric field are zero. (The same can be said for an electric dipole.)

 

[khfrekek1992]

Oh okay that makes sense, thank you! So I found the potential by taking kq/r, and now I am taking -(grad)V to get E. So I have E=-(d/dxx-hat+d/dyy-hat+d/dzz-hat)*(3Qcos(theta))/4*Pi*R^3Epsilon0

is that right?

 

[khfrekek1992]

oh wait if i take the gradient of that I get 0 in every direction..

 

[SammyS]

Oh okay that makes sense, thank you! So I found the potential by taking kq/r, and now I am taking -(grad)V to get E. So I have E=-(d/dxx-hat+d/dyy-hat+d/dzz-hat)*(3Qcos(theta))/4*Pi*R^3Epsilon0

is that right?

How did you find V ?

 

[khfrekek1992]

See below

 

[khfrekek1992]

Actually, I guess I should Multiply the charge density by the volume of the sphere to get Q, then use that in V=kQ/R to get V=(QR^2/repsilon0)cos(theta) Where everything except for r and theta is a known constant. Then I need to take the negative gradient to get E. I just don't know how to take the x,y, and z derivatives of that voltage, because they would all be 0? :/

 

[berkeman]

Yeah haha it is the same but I am doing the problem completely different now(from your suggestion) and am now stuck at a different part :/

Please do not multiple post. It is against the PF rules.

Two threads merged into one.