# Electric Field from Non-Uniformly Polarized Sphere

• Albino173
Gauss's Law and found that the potential V(r) is given by$$V(r)=-\frac {4\pi \alpha r^{n+2}+4\pi \alpha a^{n+2}}{r^2}\tag{1}$$ where α and n are constants. The potential inside the sphere is given by$$V(r)_i=-\frac {\alpha} {\epsilon_0} (n+1)(r^{n+1}-a^{n+1})$$ where n must be greater than 0.f

#### Albino173

Homework Statement
A sphere of radius 'a' has a radial polarization of ##\vec P=αr^n\hat r## where α and n are constants and n ##\geq## 0. Find volume and surface densities of bound charge. Find ##\vec E## outside and inside the sphere. Find ##V(\vec r)## (potential) outside and inside the sphere.
Relevant Equations
1) ##ρ_b=-\nabla \cdot \vec P##

2) ##σ_b=\vec P \cdot \hat n##

3) ## V(\vec r)=\frac 1 {4πε_0} \int_V \frac {ρ_b} R\, dτ + \frac 1 {4πε_0} \int_S \frac {σ_b} R\, da##

4) ##\vec E=-\nabla V(\vec r)##
Solving for the volume and surface bound charge densities was easy using equations 1) and 2).
The polarization only has an r component so
##ρ_b=-\frac 1 {r^2} \frac {d} {dr} (r^2 \vec P)=-α(n+2)r^{n-1}##,
and ##\hat n=\hat r## so
##σ_b=αa^n##.

To find ##\vec E## I intend to use equation 3) to find the potential at a general field point and then use equation 4). Since the sphere is radially symmetric, I should be able to treat the field point as on the z-axis to make the math easier.
So for the denominators of equation 4) I have
##R=(z^2+r^2-2zr\sin\theta)^{1/2}##
with 'z' being the distance from the origin to the field point and 'r' being the distance from the origin to the differential charge to be integrated. I plan to replace 'r' with 'a' in the R denominator in the surface integral term.

Plugging in equations 1), 2), 'R', and my differentials ##dτ=r^2\sin\theta dr d\theta d\phi## and ##da=r^2\sin\theta d\theta d\phi## into equation 3) I get

## V(\vec r)=\frac 1 {4πε_0} \int_0^{2\pi} \int_0^\pi \int_0^r\frac {-α(n+2)r^{n+1}\sin\theta} {(z^2+r^2-2zr\sin\theta)^{1/2}}\,dr\,d\theta\, d\phi + \frac 1 {4πε_0} \int_0^{2\pi} \int_0^\pi \frac {αa^{n+2}\sin\theta} {(z^2+a^2-2za\sin\theta)^{1/2}}\,d\theta\, d\phi## .

I already have a plan to integrate over ##\theta## by using u substitution but the volume term integral over 'r' is stumping me. I have been using the step-by-step help on integral-calculator.com to improve my integration skills but even that website got stumped with this. I'm afraid I may have set up the 'R' equation wrong but I don't know how. I'm wondering if I might have to use Guass' Law to find the ##\vec E## field since the equation for ##Q_b## doesn't involve the 'R' in the denominator. But then I would have to do the opposite of equation 4) to solve for the potential V.

• Delta2
The problem as stated suggests finding E first. No need to rederive Gauss's Law.

Homework Statement:: A sphere of radius 'a' has a radial polarization of ##\vec P=αr^n\hat r## where α and n are constants and n ##\geq## 0. Find volume and surface densities of bound charge. Find ##\vec E## outside and inside the sphere. Find ##V(\vec r)## (potential) outside and inside the sphere.
Relevant Equations:: 1) ##ρ_b=-\nabla \cdot \vec P##

2) ##σ_b=\vec P \cdot \hat n##

3) ## V(\vec r)=\frac 1 {4πε_0} \int_V \frac {ρ_b} R\, dτ + \frac 1 {4πε_0} \int_S \frac {σ_b} R\, da##

4) ##\vec E=-\nabla V(\vec r)##
I'm wondering if I might have to use Guass' Law to find the E→ field since the equation for Qb doesn't involve the 'R' in the denominator. But then I would have to do the opposite of equation 4) to solve for the potential V.
Seems to me you have to do just that. Use the integral form of Gauss's law by choosing as surface a sphere of radius r and take cases ##r\leq a## and ##r>a## . You should take advantage of the spherical symmetry of the problem in using Gauss's law in integral form. And once u know the electric field use that $$V(r)=\int_r^{\infty} \mathbf{E}\cdot d\mathbf{l}$$ where the path of integration is along the radius of the sphere. I expect ##\mathbf{E}## to be in the radial direction so the dot product that appears inside the above integral ##\mathbf{E}\cdot d\mathbf{l}## will have a nice simplified form (since we choose the path of integration along the radial direction too).

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Hint: Use the multipole expansion to solve your integrals!

• etotheipi
Hint: Use the multipole expansion to solve your integrals!
Something tells me he hasn't been taught the multipole expansion yet. The problem is kind of easy if you work with Gauss's law in integral form and find the E-field first and then the potential V.

• vanhees71
Thanks, Delta2. I guess I should have tried to work it out using Gauss' Law before even posting. For the total bound charge, I got ##Q=-4\pi \alpha r^{n+2} +4\pi \alpha a^{n+2}## which makes sense since when r=a to find the total charge, Q=0 since the sphere should be neutral.

Then to find the ##\vec E## field and potential inside the sphere I only used the first term in Q because inside the sphere the surface charge is ignored. I got
##\vec E_i=-\frac {\alpha} {\epsilon_0} r^n##.

For ##V(r)##, you showed doing the integral from r to infinity but shouldn't I integrate from 'r' to 'a' for the potential inside the sphere and then integrate from 'a' to infinity for the potential outside the sphere? That would actually be doing the same thing just split up into two integrals. I did that and got
##V(r)_i=\frac {\alpha} {\epsilon_0} (n+1)(r^{n+1}-a^{n+1})##.
I'm not 100% sure on doing this for the ##\vec E## field outside the sphere because ##Q_{total}=0## but I'm thinking that makes ##\vec E_o=0## and thus also ##V(r)_o=0##.
I'm not sure on this because I thought there could be a potential from a charge distribution even outside of where the charge is so shouldn't there be some potential outside the sphere?

I have studied the multipole expansion a bit but don't understand it enough to apply it to a problem like this. Any tips would be great.

You are mostly correct.

First one typo you got, in the formula for ##V(r)_i## the ##(n+1)## should be in the denominator.

Also to find the potential ##V(r)## we always integrate from ##r## to ##\infty##. It is just that we have to break the integral in two parts when ##r<a##, one part from ##r## to ##a##, and one from ##a## to ##\infty## , because the E-field is different inside and outside the sphere.

The total bound charge in the whole sphere is zero , as it should be, so by Gauss's law and the spherical symmetry the E-field is zero outside the sphere. That makes the potential also zero outside the sphere as you said. In this case there isn't any potential outside the sphere, because the surface bound charge density cancels out the effect of the volume bound charge density for the region external to the sphere.

• vanhees71
Then it's perhaps most easy to just integrate
$$\Delta V=-\frac{1}{\epsilon}_0 \rho$$
with
$$\rho=\rho_{\text{b}}+\sigma_{\text{b}} \delta(r-a)$$
with
$$\rho_{\text{b}}=-\vec{P}, \quad \sigma_{\text{b}}=\vec{e}_r \cdot \vec{P}|_{r=a},$$
as given in #1.