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Consider a solid insulating sphere of radius $R$ with charge density varying as $\rho = \rho _0r^2$ ($\rho _0$ is a constant and $r$ is measure from centre). Consider two points $A$ and $B$ at distance $x$ and $y$ respectively $(x < R, y > R)$ from the centre. If magnitudes of electric fields at points $A$ and $B$ are equal, then
${x^2}y = {R^3}$
${x^3}y^2 = {R^5}$
${x^2}y^3 = {R^5}$
$\frac{{{x^4}}}{y} = {R^5}$
Solution
${{\rm{E}}_{\rm{x}}} \times 4\pi {{\rm{x}}^2} = \frac{{{\rho _0}\int_0^{\rm{x}} {{{\rm{x}}^2}} \times 4\pi {{\rm{x}}^2}{\rm{dx}}}}{{{ \in _0}}}$
$ = \frac{{4\pi {\rho _0}{x^5}}}{{5{ \in _0}}}$
${{\rm{E}}_y} = \frac{{{\rho _0}\int_0^{\rm{R}} {{{\rm{x}}^2}} \times 4\pi {{\rm{x}}^2}{\rm{dx}}}}{{{ \in _{\rm{u}}} \times 4\pi {{\rm{y}}^2}}} = \frac{{{\rho _0} \times {{\rm{R}}^5}}}{{5{ \in _0}{{\rm{y}}^2}}}$
$\frac{{{\rho _0}{{\rm{x}}^3}}}{{5{ \in _0}}} = \frac{{{\rho _0}{{\rm{R}}^5}}}{{5{ \in _0}{{\rm{y}}^2}}}$
$\mathrm{x}^{3} \mathrm{y}^{2}=\mathrm{R}^{5}$
