Let a total charge 2Q be distributed in a sph | vedclass

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(C) First,we find the constant $k$ in terms of $Q$ and $R$ by integrating the charge density over the volume of the sphere:$2Q = \int_{0}^{R} \rho(r)

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$a = 8^{-1/4}R$

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(C) First,we find the constant $k$ in terms of $Q$ and $R$ by integrating the charge density over the volume of the sphere:$2Q = \int_{0}^{R} \rho(r) 4\pi r^2 dr = \int_{0}^{R} (kr) 4\pi r^2 dr = 4\pi k \int_{0}^{R} r^3 dr = 4\pi k \frac{R^4}{4} = \pi k R^4$.Thus,$k = \frac{2Q}{\pi R^4}$.Next,we find the electric field $E$ at distance $a$ from the centre using Gauss's Law:$E(4\pi a^2) = \frac{q_{enc}}{\varepsilon_0} = \frac{1}{\varepsilon_0} \int_{0}^{a} (kr) 4\pi r^2 dr = \frac{4\pi k}{\varepsilon_0} \frac{a^4}{4} = \frac{\pi k a^4}{\varepsilon_0}$.$E = \frac{k a^2}{4\varepsilon_0} = \frac{(2Q/\pi R^4) a^2}{4\varepsilon_0} = \frac{Q a^2}{2\pi \varepsilon_0 R^4}$.For charge $A$ (or $B$) to experience no force,the force due to the sphere must be balanced by the force due to the other charge:$|qE| = \frac{1}{4\pi \varepsilon_0} \frac{|Q||-Q|}{(2a)^2} \implies Q \left( \frac{Q a^2}{2\pi \varepsilon_0 R^4} \right) = \frac{Q^2}{16\pi \varepsilon_0 a^2}$.Simplifying: $\frac{a^2}{2 R^4} = \frac{1}{16 a^2} \implies a^4 = \frac{R^4}{8} \implies a = \frac{R}{8^{1/4}} = 8^{-1/4}R$.

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