# Calculate the bound charges and field in and outside a polarized sphere [closed]

The following question concerns Problem 4.10 from Griffiths Introduction to Electrodynamics.

(Boldface letters are vectors, unit vectors have a hat)

A sphere of radius $R$ carries a polarization $\mathbf {P(r)}=k\mathbf r$, where $k$ is a constant and $\mathbf r$ is the vector from the center.

a) Calculate the bound charges $\sigma_b$ and $\rho_b$.

Well I know $\sigma_b=\mathbf P \cdot \hat n$, but then my answer sheet just says $\sigma_b=kR$.

Also $\rho_b =-\nabla\cdot\mathbf P$, but then my answer sheet just says $\rho_b = -3k$.

Then for the b part

b) Find the field in and outside the sphere.

First inside:

Using a Gaussian surface I got $\mathbf E= \frac 1{3 \epsilon_0}\rho r\hat r$, the answer on my sheet is $\mathbf E= -\frac k{ \epsilon_0}\mathbf r$

I guess they just fill in the value of $\rho$ found at a), however why did the $r\hat r$ go and where did the $\mathbf r$ come from?

Now outside:

My answer sheet just says $\mathbf E=0$, how do they know this?

## closed as off-topic by sammy gerbil, ZeroTheHero, Yashas, Qmechanic♦Mar 26 '17 at 13:29

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For part a) recall that $\hat{n}$ is the unit vector perpendicular to the normal of the dielectric. In this case the dielectric is a sphere so the unit normal is $\hat{r}$ and therefore at the surface we have $\textbf{P}(R)= k R \hat{r} \implies \sigma_b = k R \hat{r}.\hat{r} = k R$. For the bound charge use cartesian co-ordinates $- (\partial_x x + \partial_y y + \partial_z z)k = -3k$
For part b) note that $\hat{r}= \frac{\textbf{r}}{r} \implies r\hat{r} = \textbf{r}$ remember we divide by the magnitude of the vector to get a unit vector. Lastly the electric field outside is zero because the sphere has a net charge of zero. Why? $-3k\frac{4 \pi R^3}{3}+ (4 \pi R^2) kR = 0$