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Suppose that you have a sphere of radius $R$ and uniform charge density $\rho$; a cylindrical hole with radius $a$ ($a\ll R$) is drilled through the center of the sphere, leaving it like a "necklace bead".

I would like to find a function for the electric field (1) very far away from the sphere ($r\gg R$) and (2) inside the hole, near the center of the bead $r\ll R$.

In case (1), I simply treat it as a point charge and calculating the electric field is trivial.

However, I am uncertain how to approach part (2) and would appreciate any assistance. The combination of spherical and cylindrical geometries seems to make this quite tricky. I am unsure what approximation or simplification to make from the knowledge that $r\ll R$.

Would it perhaps be correct to find the electric field from (1) a complete, uniformly charged sphere and (2) a cylinder of charge density $-\rho$? Summed together, the charge densities would result in our original "bead" system, so then I can just add together the expressions for the electric field. Doing case (1) is quite easy, but (2) is nontrivial for positions that are not along the axis of the cylinder, but perhaps due to our condition that $r\ll R$ and $a\ll R$, we can assume that the field from the cylinder along the $z$-axis is a good enough approximation.

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  • $\begingroup$ At the very center of the bead, symmetry implies that the electric field is zero. $\endgroup$ – Chris Mueller Apr 10 '14 at 17:13
  • $\begingroup$ Of course, but what about elsewhere in the hole? $\endgroup$ – Marcus Emilsson Apr 10 '14 at 17:14
  • $\begingroup$ The field strength should also be approximately unchanged while moving in the $\rho$ direction (in cylindrical coordinates) for the same reason that you don't feel the mass of a shell when inside of the shell. So, you should end up with something that is just a function of $z$. You're edit sounds like a good idea. It is certainly fine, in theory, because of superposition although I don't know if it will make the integrals any easier. $\endgroup$ – Chris Mueller Apr 10 '14 at 17:19
  • $\begingroup$ Don't forget that in your proposed method of adding a cylinder of charge density $-\rho$, you're either neglecting the curved portion of the sphere above and below the cylinder of negative charge or adding extra negative charges above and below the sphere, depending on the height of your cylinder. I don't know if this would be significant for the problem, but it's worth keeping in mind. (Not sure if that makes sense - if it doesn't, I'll make a picture.) $\endgroup$ – Shivam Sarodia Apr 10 '14 at 18:02
  • $\begingroup$ @Draksis: Ah, that's right. I think it's relatively safe to approximate the hole as as a cylinder, because we're given that the radius of the hole is much smaller than the radius of the sphere. $\endgroup$ – Marcus Emilsson Apr 10 '14 at 18:30
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For a cylinder:

Cylinder


$$dV=\pi a^2dr\\dq=\rho dV=\rho\pi a^2dr\\dE=Kdq/r^2=K\rho\pi a^2dr/r^2\\E=\int dE=K\rho\pi a^2\int_{r_0}^{r_0+l} dr/r^2=\frac{K\rho\pi a^2l}{r_0(r_0+l)}$$


In case inside of it as in the figure the field due to $R-x$ length cylinder is cancelled by a similiar one in the opposite side thus the resultant field is:


$$E=\frac{K\rho\pi a^2l}{r_0(r_0+l)} =\frac{K\rho\pi a^2(2x)}{(R-x)({R-x}+2x)}\\ =\frac{2K\rho\pi a^2x}{R^2-x^2} =\frac{2K\rho\pi a^2x}{R^2\left(1-\frac{x^2}{R^2}\right)} \approx\frac{2K\rho\pi a^2x}{R^2} \text{ as } x\ll R\\ =\frac{\rho a^2x}{2\epsilon_0R^2}$$


And for sphere:

$$\large E=\begin{cases} \frac{\rho x}{3\epsilon_0}\;0\le x\le R \\\frac{\rho R^3}{3\epsilon_0 x^2}\;x\ge R \end{cases}$$


Now $E$ can be easily calculated

$$E_{out}= \frac{\rho R^3}{3\epsilon_0 r^2}-\frac{\rho a^2(2R)}{4\epsilon_0(r-R)(r-R+2R)}\\ =\frac{\rho R^3}{3\epsilon_0 r^2}-\frac{2\rho a^2R}{4\epsilon_0(r^2-R^2)}\\ \approx \frac{\rho R^3}{3\epsilon_0 r^2}-\frac{\rho a^2R}{2\epsilon_0r^2} \text{ as } r\gg R \\ =\frac{\rho R}{6\epsilon_0 r^2}\left[2R^2-3a^2\right] $$

Note that, $\large\lim_{a\to 0}E=\frac{\rho R^3}{3\epsilon_0 r^2}$


Similiarly

$$E_{in}=\frac{\rho x}{3\epsilon_0}-\frac{\rho a^2x}{2\epsilon_0R^2}\\ =\frac{\rho x}{6\epsilon_0}.\left[2-3\frac{a^2}{R^2}\right]$$

Here also, $\large\lim_{a\to 0}E=\frac{\rho x}{3\epsilon_0}$


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I agree with the result, but I would like explain another more general and rapid approach. Because of the radius of the hole is negligible with respect to the radius of the sphere, and the only posible direction for E compatible with the symmetry is the z axis, and finally having in mind that the tangencial components of E are continuous, the solution is exactly the same we obtain when only considering the sphere with a uniforme charge distribution.

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