A solid ball of radius R has a charge density | vedclass

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(D) The total charge $q$ inside the ball is found by integrating the charge density over the volume of the ball:$q = \int_0^R \rho(r) \cdot 4\pi r^2 d

Vedclass · Accepted Answer

$\frac{\rho_0 R^3}{12\varepsilon_0 r^2}$

Vedclass · Answer

(D) The total charge $q$ inside the ball is found by integrating the charge density over the volume of the ball:$q = \int_0^R \rho(r) \cdot 4\pi r^2 dr$$q = \int_0^R \rho_0 \left( 1 - \frac{r}{R} \right) 4\pi r^2 dr$$q = 4\pi \rho_0 \int_0^R \left( r^2 - \frac{r^3}{R} \right) dr$$q = 4\pi \rho_0 \left[ \frac{r^3}{3} - \frac{r^4}{4R} \right]_0^R$$q = 4\pi \rho_0 \left( \frac{R^3}{3} - \frac{R^4}{4R} \right) = 4\pi \rho_0 \left( \frac{R^3}{12} \right) = \frac{\pi \rho_0 R^3}{3}$Using Gauss's Law for a point outside the ball at distance $r$ $(r > R)$:$E \cdot 4\pi r^2 = \frac{q}{\varepsilon_0}$$E \cdot 4\pi r^2 = \frac{\pi \rho_0 R^3}{3 \varepsilon_0}$$E = \frac{\rho_0 R^3}{12 \varepsilon_0 r^2}$

$A$ solid ball of radius $R$ has a charge density $\rho$ given by $\rho = \rho_0 \left( 1 - \frac{r}{R} \right)$ for $0 \leq r \leq R$. The electric field outside the ball is

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