# Flux through hollow non-conducting sphere (Question source: Halliday Resnick, 9th edition- exercise 23, Q20.)

The above question says that A and B are two non-conducting spheres enclosing a point charge in the centre. The second figure shows a graph of electric flux as a function of radius r. The scale of the vertical axis is set by $\Phi_s = 5.0 \times 10^5$ Nm$^2$/C. (a) What is the charge of the central particle? What are the net charges of (b) shell A and (c) shell B?

So to begin with, I understand that both shells would be uniformly charged since the field is uniform. However, I don't see the point of making them nonconducting shells, if we are treating them as metallic gaussian surfaces anyway unless insulators behave differently where electrostatic flux is concerned.

I don't understand why, for two values of $r$, there is a steep decline (multiple values) of $\Phi$. Could someone please explain?

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• Shell $A$ and shell $B$ have charges on them. – Farcher Apr 22 '17 at 6:55
• Yes, they would have charges on them (B -ve and A +ve), but would the flux be affected by the nonconducting material? – Antara Kulkarni Apr 22 '17 at 7:11
• If you like this is the link for solutions for resnick halliday archive.org/details/solutionstophysi029768mbp – Pranjal Rana Oct 1 '17 at 8:16

## 1 Answer

You're given net flux, and you're able to use that to solve for $\frac{q}{\epsilon_0}$, as the net flux will, by Gauss' Law, equal this value.

Gauss' Law says that the net electric flux equals the net charge enclosed divided the permittivity of free space $\epsilon_0$. So yes, the spheres have a charge on them and will affect the flux. The idea is that because A and B both have charges on them, when you draw a Gaussian surface inside A, for example, and then one between A and B, you have different enclosed charges, thus different flux. This is why there's that steep decline, because you're enclosing different charges inside and outside shell A and B.

$\epsilon_0$, by the way, is equal to about $8.854*10^{-12}F/m$.

So inside, when $r<A$, $2*10^5 = \frac{q}{\epsilon_0}$, so you can solve and get the charge is $1.77*10^{-6} C$.

You can follow this same procedure but now knowing that the value of the charge, when you solve for $q$ when $A < r < B$, in between the shells, you'll get the total charge enclosed. You will then find the difference between the charge enclosed and the point charge, and that will be the charge on shell $A$. Same logic applies to $B$.