Answer to Question #174832 in Electricity and Magnetism for Robert Cunningham

Question

A sphere of charge Q and radius R has an electric field 𝐸(𝑟) = 𝑟4𝐸𝑚𝑎𝑥/𝑅4 for r ≤ R. (a) Find the charge density of the sphere as a function of r. (b) Find the potential function in all space for this object. Give your answers in terms of Q, R, and r.

Answer 1

Given,

$𝐸(𝑟) = 𝑟^4\dfrac{𝐸_{𝑚𝑎𝑥}}{𝑅^4}$

As per the Gauss rule,

$\oint E.ds=\dfrac{q_{in}}{\epsilon_o}$

$\Rightarrow E\oint ds=\dfrac{q_{in}}{\epsilon}$

$\Rightarrow E (4\pi r^2)=\dfrac{q_{in}}{\epsilon}$

$\Rightarrow E = \dfrac{q_{in}}{4\pi \epsilon_o r^2}$

if r=R,

$E_{max}=\dfrac{Q}{4\pi \epsilon_o R^2}$

As per the Gauss rule,

$\oint E(r).ds=\dfrac{q_{in}}{\epsilon_o}$

$\Rightarrow 𝑟^4\dfrac{𝐸_{𝑚𝑎𝑥}}{𝑅^4}. 4\pi r^2=\dfrac{q_{in}}{\epsilon_o}$

$\Rightarrow \dfrac{Q}{4\pi \epsilon_o R^2}\dfrac{4\pi r^6}{R^4}=\frac{q_{in}}{\epsilon_o}$

$\Rightarrow \dfrac{Qr^6}{R^6}={q_{in}}$

We know that $dq=\rho dV$

we know that $dV=4\pi r^2 dr$

$\int dq=\int\rho(r)4\pi r^2 dr =q_{in}$

$\Rightarrow \int\rho(r)4\pi r^2 dr =q_{in}$

$\Rightarrow \int\rho(r) r^2 dr =\dfrac{Q r^6}{4\pi R^6}$

Now, taking the differentiation,

$\Rightarrow \frac{d}{dr}(\int\rho(r) r^2 dr) =\dfrac{Q }{4\pi R^6}\frac{d(r^6)}{dr}$

$\Rightarrow {\rho(r) r^2}=\dfrac{6Qr^5}{4\pi R^6}$

$\Rightarrow \rho(r)=\dfrac{3 Qr^3}{2\pi R^6}$

For potential function,

$\int dV= \int \dfrac{{dq}}{4\pi \epsilon_o r}$

$dq=\rho(r)(4\pi r^2)dr$

$\Rightarrow V = \frac{}{}$ $\int \dfrac{\rho(r)(4\pi r^2)dr}{4\pi \epsilon_o r}=\int\dfrac{3 Qr^3}{\epsilon_o2\pi R^6}( r)dr$

$\Rightarrow V = \dfrac{3Q}{2\pi \epsilon_o R^6}\int(r^3)dr$

$V=\dfrac{3Qr^4}{8\pi \epsilon_oR^6}$