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(A) The electric field at point $P$ is the vector sum of the electric field due to the solid cylinder (without the cavity) and the electric field due

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$6$

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(A) The electric field at point $P$ is the vector sum of the electric field due to the solid cylinder (without the cavity) and the electric field due to a sphere of radius $R/2$ with charge density $-\rho$.$1$. Electric field due to the solid cylinder at distance $r = 2R$:Using Gauss's Law,$E_1 = \frac{\lambda}{2 \pi \varepsilon_0 r}$,where $\lambda = \rho \pi R^2$.$E_1 = \frac{\rho \pi R^2}{2 \pi \varepsilon_0 (2R)} = \frac{\rho R}{4 \varepsilon_0}$.$2$. Electric field due to the spherical cavity (treated as a sphere of charge density $-\rho$):The charge of the sphere is $q = -\rho \cdot \frac{4}{3} \pi (R/2)^3 = -\rho \cdot \frac{4}{3} \pi \cdot \frac{R^3}{8} = -\frac{\rho \pi R^3}{6}$.The electric field at distance $2R$ from the center is $E_2 = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{|q|}{(2R)^2} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{\rho \pi R^3 / 6}{4R^2} = \frac{\rho R}{96 \varepsilon_0}$.$3$. Net electric field $E = E_1 - E_2$:$E = \frac{\rho R}{4 \varepsilon_0} - \frac{\rho R}{96 \varepsilon_0} = \frac{\rho R}{\varepsilon_0} \left( \frac{24 - 1}{96} \right) = \frac{23 \rho R}{96 \varepsilon_0}$.Given $E = \frac{23 \rho R}{16 k \varepsilon_0}$,we have $\frac{23 \rho R}{96 \varepsilon_0} = \frac{23 \rho R}{16 k \varepsilon_0}$.Thus,$96 = 16k \Rightarrow k = 6$.

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