- #1

Albino173

- 5

- 1

- 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.

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.