From Griffiths, Problem 4.1(adsbygoogle = window.adsbygoogle || []).push({});

A sphere of radius R carries a polarization

[tex]

\textbf{P}=k\textbf{r}

[/tex]

where k is constant and r is the vector from the center.

a. Calculate [tex]\sigma_b[/tex] and [tex]\rho_b[/tex].

b. Find the field inside and outside the sphere.

part a is handled simply by [tex]\sigma_b=\textbf{P}\cdot\hat{n}[/tex] and [tex]\rho_b=-\nabla\cdot\textbf{P}[/tex].

part b is handled most easily by using the bound charges found and gauss's law, giving: [tex]\textbf{E}=\frac{-kr}{\epsilon_0}\hat{r}[/tex] and 0 outside.

part b can also be handled by first getting the potential through:[tex]\frac{1}{4\pi \epsilon_0}\int_v \frac{\hat{r}\cdot\textbf{P}}{r^2}d\tau[/tex]. Then just get the negative gradient of V to yield E. This gives the same result as method 1 for both inside and outside (P=0 when dealing with outside case).

Now, i tried to get the field, again by computing for V but this time with:

[tex]V(r)=\frac{1}{4\pi \epsilon_0}\oint_s\frac{\sigma_b}{r}da + \frac{1}{4\pi \epsilon_0}\int_v\frac{\rho_b}{r}d\tau[/tex]. The answer i get using this integral is different. i would like to know if in the first place this integral is applicable? if i'm trying to get the potential inside, do i still consider the surface bound charge?

the best i managed was an answer negative to the correct one. (but this had 1 dubioius step involved). also, i could only attempt to get the potential inside. can anyone make the last method work (for both outside and inside)? or share why it doesnt?

thanks.

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