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Question

(C) Consider a spherical shell of radius $x$ and thickness $dx$. The charge on this shell is given by $dq = \rho(x) \cdot 4 \pi x^2 dx = \rho_0 \left(

Vedclass · Accepted Answer

$\frac{\rho _0 r}{4 \varepsilon _0} \left( \frac{5}{3} - \frac{r}{R} \right)$

Vedclass · Answer

(C) Consider a spherical shell of radius $x$ and thickness $dx$. The charge on this shell is given by $dq = \rho(x) \cdot 4 \pi x^2 dx = \rho_0 \left( \frac{5}{4} - \frac{x}{R} \right) \cdot 4 \pi x^2 dx$.The total charge $q$ enclosed within a sphere of radius $r$ $(r $q = \int_0^r dq = 4 \pi \rho_0 \int_0^r \left( \frac{5}{4} x^2 - \frac{x^3}{R} \right) dx$$q = 4 \pi \rho_0 \left[ \frac{5}{4} \cdot \frac{r^3}{3} - \frac{1}{R} \cdot \frac{r^4}{4} \right] = 4 \pi \rho_0 \left( \frac{5 r^3}{12} - \frac{r^4}{4R} \right) = \pi \rho_0 r^3 \left( \frac{5}{3} - \frac{r}{R} \right)$.Using Gauss's Law,the electric field $E$ at distance $r$ is:$E \cdot 4 \pi r^2 = \frac{q}{\varepsilon_0}$$E = \frac{1}{4 \pi \varepsilon_0 r^2} \cdot \left[ \pi \rho_0 r^3 \left( \frac{5}{3} - \frac{r}{R} \right) \right]$$E = \frac{\rho_0 r}{4 \varepsilon_0} \left( \frac{5}{3} - \frac{r}{R} \right)$.

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