Physics Friday 36

Consider a ball of radius R and uniform electric charge density ρ. What is the electric field at a point a distance r from the center of the ball? Suppose we remove material from this ball to form a smaller spherical cavity within it, such that the vector from the center of the overall sphere to that of the cavity is d. What, then, is the electric field at any point within the cavity?

I. To calculate the field, we first consider the spherical symmetry. The magnitude of the field will be a function of r only (in terms of the point’s location), and the field will also be in a radial direction. So, let us consider as a Gaussian surface a sphere of radius r with the same center as the ball. Then, as the field is of constant magnitude across that surface, and normal to the sphere at each point, the total flux of the electric field through the surface is just the magnitude E of the field times the surface area: Φ=4πr²E. Now, Gauss’ Law tells us that this flux is , where Q is the total charge enclosed in the sphere.

For r>R, this is the total charge of the ball: . Thus
,
which is the same as the field at a distance r from a point charge .

For r<R, the enclosed charge is instead $Q=\frac{4}{3}\pi{r^3}\rho$ . Then,
$\begin{eqnarray}E&=&\frac{\text{\Phi}}{4\pi{r^2}}\\&=&\frac{Q}{4\pi{r^2}\epsilon_0}\\&=&\frac{\rho}{3\epsilon_0}r\end{eqnarray}$ .

For both of these, the field is radial (outward for ρ>0, inward for ρ<0).

II. Note, that if we place a ball of charge density ρ and radius R and a smaller sphere of charge density -ρ with center displaced from the first ball by the vector d, the result is equivalent, electrically, to the ball with cavity, due to the cancellation of charge.

Thus, a point within the cavity will be, in this equivalent picture, within both balls. Let the vector from the center of the larger ball to our point in question be r. Then the vector from the center of our smaller ball (aka the cavity) to this point is r-d.

The field, then, from the larger ball is of magnitude $E_1=\frac{\rho}{3\epsilon_0}r$ , where r is the magnitude of the vector r. This field is radially directed to the sphere, and thus $\mathbf{E}_1=\frac{\rho}{3\epsilon_0}\mathbf{r}$ .

Similarly, for field due to the smaller sphere, $E_2=\frac{-\rho}{3\epsilon_0}|\mathbf{r}-\mathbf{d}|$ , and so
$\mathbf{E}_2=\frac{-\rho}{3\epsilon_0}(\mathbf{r}-\mathbf{d})=\frac{\rho}{3\epsilon_0}(\mathbf{d}-\mathbf{r})$ .

Summing these:
$\begin{eqnarray}\mathbf{E}&=&\mathbf{E}_1+\mathbf{E}_2\\&=&\frac{\rho}{3\epsilon_0}\mathbf{r}+\frac{\rho}{3\epsilon_0}(\mathbf{d}-\mathbf{r})\\&=&\frac{\rho}{3\epsilon_0}\mathbf{d}\end{eqnarray}$ .

So the field inside the cavity is uniform, and independent of the size of the cavity. Note that if the cavity is concentric to the ball (d=0), the field is zero.

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Tags: Electricity & Magnetism, Electrostatics, Friday Physics, Gauss' Law, physics, symmetry

This entry was posted on September 5, 2008 at 12:01 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

One Response to “Physics Friday 36”

Physics Friday 121 « Twisted One 151's Weblog Says:
May 31, 2010 at 3:24 pm | Reply
[...] charge densities rather than discrete points, the more useful form of this is Gauss’ Law (see here and here for previous uses). In integral form, Gauss’ Law reads: . Using the divergence [...]

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