A uniformly charged sphere of charge density ρ_0 and radius R is surrounded by a charged medium of volume charge density ρ=α/r where α is a positive constant and r is the distance from the center of the sphere. I think it might be solvable using the uniqueness theorem/method of images but I don't know how to go about it.


closed as off-topic by Rob Jeffries, Yashas, John Rennie, Kyle Kanos, Bill N May 10 '17 at 13:42

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  • $\begingroup$ Just use the Gauss' theorem IMO. $\endgroup$ – Abhijeet Melkani May 10 '17 at 6:15
  • $\begingroup$ Using Gauss law gave me a ρ_0 that increases as the square of r which is not possible ( ρ_0 is supposed to be constant) $\endgroup$ – Derby Moose May 10 '17 at 6:22
  • $\begingroup$ Unless of course ρ_0 is equal to 0. But I don't think the answer is that simple( The question talks about a "charged" sphere) $\endgroup$ – Derby Moose May 10 '17 at 6:23
  • $\begingroup$ Disregard my previous comments, I made an error in the integral. $\endgroup$ – Derby Moose May 10 '17 at 7:51
  • $\begingroup$ Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. $\endgroup$ – Kyle Kanos May 10 '17 at 10:07

Let $dq$ be the charge enclosed in a spherical shell of radius $r$ and thickness $dr$ concentric with the given sphere. For $r>R$ we have:

$$dq=4\pi r^2\rho dr\\ dq=4\pi r^2\frac{\alpha}{r} dr\\dq=4\pi \alpha r dr$$

the total charge $Q$ enclosed in a sphere of radius $x$ will be

$$Q=\frac43\pi R^3 \rho_0+\int_R^x4\pi\alpha rdr\\Q=\frac43\pi R^3 \rho_0 +2\pi\alpha(x^2-R^2) $$

Applying Gauss' law we get the magnitude $E$ of te electric field at $x$

$$ E=\frac{Q}{4\pi x^2 \epsilon_0}\\E=\frac{\frac43\pi R^3 \rho_0 +2\pi\alpha(x^2-R^2)}{4\pi x^2 \epsilon_0}\\E=\frac{2\pi R^2(\frac23R\rho_0-\alpha)}{4\pi x^2 \epsilon_0}+\frac{\alpha}{2\epsilon_0}$$ for the dependence on $x$ to vanish, the first term must be zero. Therefore

$$\frac{2\pi R^2(\frac23R\rho_0-\alpha)}{4\pi x^2 \epsilon_0}=0\\(\frac23R\rho_0-\alpha)=0\\\rho_0=\frac{3\alpha}{2R}$$

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    $\begingroup$ Please don't answer lazy do my homework for me type questions; these types of questions are considered off topic & answering them only encourages other users to post their own lazy questions. $\endgroup$ – Kyle Kanos May 10 '17 at 10:07

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