$R=\frac{\rho l}{A}$
$\therefore \quad A=\frac{\rho l}{R}$
$\Rightarrow \frac{A_{1}}{A_{2}}=\frac{\rho_{1}}{\rho_{2}} \times \frac{L_{1}}{I_{2}}\left(\frac{R_{2}}{R_{1}}\right)$
$\Rightarrow \frac{A_{1}}{A_{2}}=1 \quad\left[\because R_{1}=R_{2}, I_{1}=l_{2}\right.$ and for same
material $\left.\rho_{1}=\rho_{2}\right]$
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$\rho (r)\, = \,{\rho _0}\left( {1 - \frac{r}{R}} \right)$ for $r < R$
$\rho (r)\,=\,0$ for $r\, \ge \,R$
Where $r$ is the distance from the centre of the charge distribution $\rho _0$ is a constant. The electric field at an internal point $(r < R)$ is