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(B) For a point $A$ inside the sphere $(x < R)$,using Gauss's Law: $E_A \cdot 4\pi x^2 = \frac{q_{enclosed}}{\epsilon_0} = \frac{1}{\epsilon_0} \int_0

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

Vedclass · Answer

(B) For a point $A$ inside the sphere $(x Substituting $\rho = \rho_0 r^2$: $E_A \cdot 4\pi x^2 = \frac{4\pi \rho_0}{\epsilon_0} \int_0^x r^4 dr = \frac{4\pi \rho_0 x^5}{5\epsilon_0}$.Thus,$E_A = \frac{\rho_0 x^3}{5\epsilon_0}$.For a point $B$ outside the sphere $(y > R)$,the sphere acts as a point charge at the centre: $E_B = \frac{Q_{total}}{4\pi \epsilon_0 y^2}$.$Q_{total} = \int_0^R \rho_0 r^2 \cdot 4\pi r^2 dr = 4\pi \rho_0 \int_0^R r^4 dr = \frac{4\pi \rho_0 R^5}{5}$.So,$E_B = \frac{4\pi \rho_0 R^5}{5 \cdot 4\pi \epsilon_0 y^2} = \frac{\rho_0 R^5}{5\epsilon_0 y^2}$.Given $E_A = E_B$,we have $\frac{\rho_0 x^3}{5\epsilon_0} = \frac{\rho_0 R^5}{5\epsilon_0 y^2}$.Simplifying,we get $x^3 y^2 = R^5$.

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