A spherically symmetric charge distribution i | vedclass

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Question

(B) According to Gauss's Law,the electric field $E$ at a distance $r$ from the centre of a spherically symmetric charge distribution is given by $E \c

Vedclass · Accepted Answer

$\frac{\rho_0}{\varepsilon_0} \left( \frac{r}{3} - \frac{r^2}{4R} \right)$

Vedclass · Answer

(B) According to Gauss's Law,the electric field $E$ at a distance $r$ from the centre of a spherically symmetric charge distribution is given by $E \cdot (4\pi r^2) = \frac{q_{enclosed}}{\varepsilon_0}$.To find the enclosed charge $q$ within a sphere of radius $r $q = \int_0^r \rho(x) \cdot 4\pi x^2 dx$Substituting $\rho(x) = \rho_0 \left( 1 - \frac{x}{R} \right)$:$q = 4\pi \rho_0 \int_0^r \left( x^2 - \frac{x^3}{R} \right) dx$$q = 4\pi \rho_0 \left[ \frac{x^3}{3} - \frac{x^4}{4R} \right]_0^r = 4\pi \rho_0 \left( \frac{r^3}{3} - \frac{r^4}{4R} \right)$Now,using Gauss's Law:$E \cdot 4\pi r^2 = \frac{4\pi \rho_0}{\varepsilon_0} \left( \frac{r^3}{3} - \frac{r^4}{4R} \right)$Dividing both sides by $4\pi r^2$:$E = \frac{\rho_0}{\varepsilon_0} \left( \frac{r}{3} - \frac{r^2}{4R} \right)$

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