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(A) To find the electric field at point $P$,we use the principle of superposition. We consider the cylinder with the cavity as the sum of a complete s

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$6$

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(A) To find the electric field at point $P$,we use the principle of superposition. We consider the cylinder with the cavity as the sum of a complete solid cylinder with charge density $\rho$ and a sphere with charge density $-\rho$.$1$. Electric field due to the infinite solid cylinder of radius $R$ at distance $r = 2R$ from the axis:Using Gauss's Law,$E_{cyl} \cdot (2\pi r L) = \frac{q_{enclosed}}{\varepsilon_0} = \frac{\rho \cdot \pi R^2 L}{\varepsilon_0}$.$E_{cyl} = \frac{\rho R^2}{2 \varepsilon_0 r} = \frac{\rho R^2}{2 \varepsilon_0 (2R)} = \frac{\rho R}{4 \varepsilon_0}$.$2$. Electric field due to the spherical cavity of radius $a = R/2$ at distance $r = 2R$ from its center:Using the formula for the electric field outside a uniformly charged sphere,$E_{sph} = \frac{kQ}{r^2} = \frac{1}{4\pi \varepsilon_0} \cdot \frac{\rho \cdot (4/3)\pi a^3}{r^2}$.Substituting $a = R/2$ and $r = 2R$:$E_{sph} = \frac{\rho \cdot (4/3)\pi (R/2)^3}{4\pi \varepsilon_0 (2R)^2} = \frac{\rho \cdot (4/3) (R^3/8)}{4 \varepsilon_0 (4R^2)} = \frac{\rho R^3 / 6}{16 \varepsilon_0 R^2} = \frac{\rho R}{96 \varepsilon_0}$.$3$. Net electric field at $P$:Since the sphere has charge density $-\rho$,the field is in the opposite direction to the cylinder's field.$E_{net} = E_{cyl} - E_{sph} = \frac{\rho R}{4 \varepsilon_0} - \frac{\rho R}{96 \varepsilon_0} = \frac{24\rho R - \rho R}{96 \varepsilon_0} = \frac{23\rho R}{96 \varepsilon_0}$.Comparing this with the given expression $\frac{23\rho R}{16K\varepsilon_0}$:$16K = 96 \implies K = 6$.

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