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

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

(b) Find the field inside and outside the sphere.

For $r>R$ I calculated the electric field to be zero and for $r<R$, I calculated the electric field to be $E = -\frac{kr}{\epsilon_0} \hat{r}$. I was reading the next chapter of Griffiths E&M about electric displacement and I went back to this problem to find the electric field using this method and check the results. The electric displacement is $D = \epsilon_0 E + P$ and \begin{equation} \int D \cdot da = (Q_f)_{\text{enc}} \end{equation} Since there are no free charges in this problem, we have that $D=0$, so \begin{equation} D = \epsilon_0 E + P=0 \;\;\;\; \implies E = -\frac{P}{\epsilon_0} \end{equation} This works out for the case $r<R$. However, for $r>R$, we should get $E=0$, which implies that $P=0$ for $r>R$. But the polarization is defined as $P(r) = kr$ which is not zero for $r>R$. My question is how would one intuitively deduce this without calculating the electric field first? For this example, I think it is because the volume density and surface charge density cancel each other out. Since $\sigma = kR $ and $\rho = -3k$ \begin{equation} Q_{\text{total}} = \sigma A + \rho V = 4\pi k R^3 -3k\left(\frac{4}{3}\pi R^3\right) = 0 \end{equation} But I have not been able to find much about it online so I am not sure of this logic. In more complicated cases how would I be able to deduce the polarization is zero in some regions?

  • $\begingroup$ I have objection to the conclusion that $\mathbf{D} =0$. (Inside the sphere) See this : physics.stackexchange.com/questions/468061/… $\endgroup$ Mar 22, 2023 at 12:20
  • $\begingroup$ I feel that in Griffiths exercise 4.10 solution, as you have written, the electric field only due to Bound Charge density is considered. If we were to include free charge density as well, then $\mathbf{P} = \varepsilon_0 \chi_e \mathbf{E} $ $\endgroup$ Mar 22, 2023 at 13:11

1 Answer 1


Assuming the sphere is sitting in vacuum, the polarization outside of the sphere is necessarily zero because there is no material outside of the sphere to generate additional dipole moments. Because of this, it's implied that the polarization $P$ is only defined up to $r = R$.

Moreover, as you point out, the surface charge must cancel out the volume charge. This is intuitive because the polarization is the dipole moment per unit volume, so the total charges due to polarization must be zero. We can show this mathematically by invoking divergence theorem on the definition of bound volume charges:

$\int \rho_b dV = \int -\nabla \cdot \mathbf{P} dV = - \int P \cdot \hat{\mathbf{n}} dS = - \int \sigma_b dS$

where $\rho_b$ and $\sigma_b$ refer to the the bound volume and surface charges, respectively. Thus, the amount of volume charge exactly cancels out the surface charge.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.