Charged enclosed by the sphere

Question

I'm reviewing the book "Conquering the Physics GRE" for my upcoming Physics GRE. I came across this problem which I'm having trouble with understanding. In particular, I understand the solution that the author provides but I don't understand what is wrong with my approach.

Q. The Electric field inside a sphere of radius $R$ is given by $E = E_0 z^2 \hat{\textbf{z}}$. What is the total charge of the sphere?

The authors approach involving taking the divergence of the electric field to get the charge density and then integrating the density over the volume of the sphere to get charged enclosed, which in their case turns out to be $0$.

But we can also just use a concentric sphere of radius $r$ ($0 < r \le R$) as a Gaussian surface and just use the integral form of Maxwell's equation to calculate the charge enclosed.

$$ \oint \limits_{S} \vec{E} \cdot d\vec{S} = \frac{Q_{enc}}{\epsilon_0} .$$

Since the area vector points in the radial direction, if we assume it makes an angle $\theta$ with the Electric Field vector, and given $z = r cos(\theta)$, we have

$$ Q_{enc} = \epsilon_0 \int \limits_{0}^{\pi} \int \limits_{0}^{2\pi} E_0 r^2 cos^2(\theta) r^2 sin(\theta) d\theta d\phi,$$

$$ Q_{enc} = \frac{4 \pi \epsilon_0 E_0}{3} r^4 . $$

If we want the charge enclosed by the sphere, we just set $r = R$, so we get

$$ Q = \frac{4 \pi \epsilon_0 E_0}{3} R^4 .$$

which isn't zero.

I'm having trouble figuring out where I'm going wrong. Any suggestions appreciated.

Answer 1

I think you forgot to account for $\mathbf{\hat{z}}$

Let $\mathbf{\hat{r}}$ be the normal to the surface of our sphere. If you take the route of integrating the electric field over the surface of the sphere that contains the charge, then you will be evaluating the following quantity.

$z^2 \mathbf{\hat{z}}.\mathbf{\hat{r}}=z^2 \mathbf{\hat{z}}.\mathbf{r}/r=z^2\, z/r=z^3/r$

So you will be integrating $z^3$ over the surface of the sphere centered on the origin. Since $z^3$ is an odd function the integral will vanish.

Answer 2

There is also a nice geometrical argument for this. Since the field is $\vec E=z^2\hat z$, the fields lines always point along $+\hat z$ and the magnitude of the field does not depend on the position in the $xy$-plane. As a result, every field line that enters the sphere must also exit the sphere, so the net flux must be $0$, and therefore the net enclosed charge must be $0$.