Consider a solid insulating sphere of radius | vedclass

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Question

(B) For a point $A$ inside the sphere $(x < R)$,the electric field $E_x$ is given by Gauss's Law:$E_x \cdot 4\pi x^2 = \frac{q_{enclosed}}{\epsilon_0}

Vedclass · Accepted Answer

$x^3 y^2 = R^5$

Vedclass · Answer

(B) For a point $A$ inside the sphere $(x $E_x \cdot 4\pi x^2 = \frac{q_{enclosed}}{\epsilon_0} = \frac{1}{\epsilon_0} \int_0^x (\rho_0 r^2) \cdot 4\pi r^2 dr$$E_x \cdot 4\pi x^2 = \frac{4\pi \rho_0}{\epsilon_0} \int_0^x r^4 dr = \frac{4\pi \rho_0}{\epsilon_0} \cdot \frac{x^5}{5}$$E_x = \frac{\rho_0 x^3}{5\epsilon_0}$For a point $B$ outside the sphere $(y > R)$,the electric field $E_y$ is given by:$E_y = \frac{Q_{total}}{4\pi \epsilon_0 y^2}$,where $Q_{total} = \int_0^R (\rho_0 r^2) \cdot 4\pi r^2 dr = \frac{4\pi \rho_0 R^5}{5}$$E_y = \frac{4\pi \rho_0 R^5}{5 \cdot 4\pi \epsilon_0 y^2} = \frac{\rho_0 R^5}{5\epsilon_0 y^2}$Given $E_x = E_y$,we have:$\frac{\rho_0 x^3}{5\epsilon_0} = \frac{\rho_0 R^5}{5\epsilon_0 y^2}$$x^3 = \frac{R^5}{y^2} \implies x^3 y^2 = R^5$

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