Electric field of a charged dielectric sphere

[1v1Dota2RightMeow]

Note: $E_{in}=-\hat{a_r} A cos(\theta)+\hat{a_{\theta}} A sin(\theta)$ and $E_{out}=\hat{a_r} 2 Bcos(\theta)/r^3 +\hat{a_{\theta}} B sin(\theta)/r^3$. (At least that's what I computed). The $\hat{a}_{\theta}$ term in the formula for the gradient in spherical coordinates has a (1/r) in it.

Wait I got something different. $E_{in} = -A (cos \theta (-\hat a_r) +sin \theta \hat a_{\theta})$ because in finding $E$ you take the negative of the gradient of $V$.

 

[Charles Link]

Wait I got something different. $E_{in} = -A (cos \theta (-\hat a_r) +sin \theta \hat a_{\theta})$ because in finding $E$ you take the negative of the gradient of $V$.

$V_{in}=Arcos(\theta)$ . This gives $E_{in}= -\nabla V_{in}=-Acos(\theta) \hat{a}_r +A sin(\theta) \hat{a}_{\theta}$. I do believe I have it correct. $\\$ One minor item of interest, but something you might find of interest is you had asked about a boundary condition involving $E$. Since $V_{in}=V_{out}$ everywhere at r=R, we can move parallel to the surface and come to a new location where again $V_{in}=V_{out}$. This means that the $\hat{a}_{\theta}$ components of E (just inside and just outside the sphere) will necessarily be equal at r=R. (Only the $\hat{a}_{\theta}$ components. ) Employing this requirement gives us the same thing as $V_{in}=V_{out}$: $A=B/R^3$. It's a somewhat minor detail, but you might find it of interest...

 

[1v1Dota2RightMeow]

$V_{in}=Arcos(\theta)$ . This gives $E_{in}= -\nabla V_{in}=-Acos(\theta) \hat{a}_r +A sin(\theta) \hat{a}_{\theta}$. I do believe I have it correct. $\\$ One minor item of interest, but something you might find of interest is you had asked about a boundary condition involving $E$. Since $V_{in}=V_{out}$ everywhere at r=R, we can move parallel to the surface and come to a new location where again $V_{in}=V_{out}$. This means that the $\hat{a}_{\theta}$ components of E (just inside and just outside the sphere) will necessarily be equal at r=R. (Only the $\hat{a}_{\theta}$ components. ) Employing this requirement gives us the same thing as $V_{in}=V_{out}$: $A=B/R^3$. It's a somewhat minor detail, but you might find it of interest...

Ok I see my mistake.

So I finished it and this is my final answer:

$V_{in}(r, \theta) = -\frac{\sigma_0 rcos \theta}{(2\epsilon_0 + \epsilon)}$ and

$V_{out}(r, \theta) = -\frac{-r \sigma_0}{(2 \epsilon_0 + \epsilon)}$.

The electric field inside is then $E_{in}=\frac{\sigma_0 ( cos \theta (-\hat a_r) - sin \theta \hat a_{\theta})}{2\epsilon_0 + \epsilon}$

Is that what you got?

 

[Charles Link]

Ok I see my mistake.

So I finished it and this is my final answer:

$V_{in}(r, \theta) = -\frac{\sigma_0 rcos \theta}{(2\epsilon_0 + \epsilon)}$ and

$V_{out}(r, \theta) = -\frac{-r \sigma_0}{(2 \epsilon_0 + \epsilon)}$.

The electric field inside is then $E_{in}=\frac{\sigma_0 ( cos \theta (-\hat a_r) - sin \theta \hat a_{\theta})}{2\epsilon_0 + \epsilon}$

Is that what you got?

I did get the $2 \epsilon_o+\epsilon$ in the denominator. I think you have a couple of corrections to make though. I got $A=\sigma_o/(2 \epsilon_o+\epsilon)$ and $B=AR^3$. The form of the $V_{out} =Bcos(\theta)/r^2$ will have a $cos(\theta)$ in the numerator and an $r^2$ in the denominator. Also I got "+" signs on both potential terms. It looks like a couple algebraic corrections will give you the answer... Also, notice the $E_{in}$ is actually uniform and points in the (minus) z-direction. A little vector algebra will show this. $\\$ You need to get the correct sign on $V_{in}$. (Note for $V_{in}$, the positive part of the free charge distribution is in the forward +z direction so that the potential will be positive at r=R for $\theta=0$.) Possibly the wrong sign on $V_{in}$ came from the wrong sign on $E=-\nabla V_{in}$ that was used in the boundary equation calculation. $\\$ Additional item: We can also compare this result to the $E_{in}$ that I computed in post #2 above. Just one simple way to show you that you did get the correct (-1/3) factor for a sphere: If you let $\epsilon=\epsilon_o$, you do get that $E_{in}=-(\sigma_o/(3 \epsilon_o)) \hat{z}$. It also agrees completely with the result in post #2 for $\epsilon=\epsilon_o (1+\chi)$ as a little additional algebra will show. $\\$ Additional comment: The couple of these Legendre problems that have appeared on Physics Forums, including yours, have been very good practice problems for solving. With a little practice, this method is starting to get a little easier.

 

[Charles Link]

And a follow-on: To summarize the results of post #2 above:
$E_i=E_o/(1+\chi/3)$ $\\$ $E_o=-((\sigma_o)/(3 \epsilon_o)) \hat{z}$ $\\$ $\epsilon=\epsilon_o (1+\chi)$ $\\$ A little algebra gives $E_i=-((\sigma_o/(2 \epsilon_o+\epsilon)) \hat{z}$ in complete agreement with the Legendre result.

 

[Pure motion]

One aspect involved in a dielectric sphere is the medium used in the dielectric, because of that... spacing between two conductors that are on each side of the dielectric sphere are critical, the dielectric is a transmitter of electric force, ( voltage ) between the two conductors... Placing a charge on the inside of a dielectric sphere would require a conductor that is less than the size of the dielectric sphere it's self . The smaller conductor inside the dielectric sphere would allow a different charge potential as compared to the conductor that is on the outside of the dielectric sphere.

 

[Charles Link]

@1v1Dota2RightMeow Please see my posts #34 and #35 above.

 

[1v1Dota2RightMeow]

@1v1Dota2RightMeow Please see my posts #34 and #35 above.

Ah, darn! I thought I was right! I already turned in the assignment, but once my professor hands it back I will review where I made a mistake!