# Electric Field from Non-Uniformly Polarized Sphere

• Albino173
In summary: 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.
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.

Albino173 said:
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).

Last edited:
Hint: Use the multipole expansion to solve your integrals!

etotheipi
vanhees71 said:
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.

## 1. What is an electric field?

An electric field is a physical quantity that describes the influence of an electric charge on other charges in its vicinity. It is represented by a vector that indicates the direction and strength of the force that a charge would experience if placed in the field.

## 2. How is an electric field created by a non-uniformly polarized sphere?

A non-uniformly polarized sphere is one in which the electric charge is not evenly distributed. This creates an uneven distribution of electric field lines, with a higher concentration of lines near areas of higher charge density. The overall electric field is the vector sum of all the individual electric fields created by each charge on the sphere.

## 3. How does the electric field from a non-uniformly polarized sphere differ from that of a uniformly polarized sphere?

In a uniformly polarized sphere, the electric charge is evenly distributed, creating a uniform electric field. In a non-uniformly polarized sphere, the uneven distribution of charge creates a non-uniform electric field, with different strengths and directions at different points in space.

## 4. What factors affect the strength of the electric field from a non-uniformly polarized sphere?

The strength of the electric field from a non-uniformly polarized sphere depends on the magnitude and distribution of the electric charge on the sphere, as well as the distance from the sphere. Additionally, the dielectric constant of the surrounding medium can also affect the strength of the electric field.

## 5. How can the electric field from a non-uniformly polarized sphere be calculated?

The electric field from a non-uniformly polarized sphere can be calculated using Coulomb's law, which states that the electric field at a point is equal to the force exerted by a point charge at that point, divided by the magnitude of the charge. Additionally, the electric field can also be calculated using Gauss's law, which relates the electric field to the charge enclosed by a surface surrounding the sphere.

Replies
1
Views
839
Replies
2
Views
868
Replies
12
Views
523
Replies
10
Views
2K
Replies
6
Views
3K
Replies
2
Views
427
Replies
7
Views
1K
Replies
26
Views
4K