Electric Field due to Cylinder with Spherical Cavity

To find the value of $k$, we use the principle of superposition. The electric field at point $P$ can be calculated by subtracting the electric field of the removed spherical cavity (filled with charge density $\rho$) from the electric field of a complete, solid infinitely long cylinder of charge density $\rho$.

Step 1: Electric Field due to the Complete Solid Cylinder

Consider an infinitely long solid cylinder of radius $R$ and uniform volume charge density $\rho$. Point $P$ is located at a distance $r = 2R$ from the axis of the cylinder.

Using Gauss's Law for a coaxial cylindrical Gaussian surface of radius $r = 2R$ and length $L$: $\oint \mathbf{E}_{\text{cylinder}} \cdot d\mathbf{A} = \frac{q_{\text{enclosed}}}{\varepsilon_0}$

The enclosed charge in length $L$ is: $q_{\text{enclosed}} = \rho \cdot (\pi R^2 L)$

Thus, $E_{\text{cylinder}} \cdot (2\pi r L) = \frac{\rho \pi R^2 L}{\varepsilon_0}$

$E_{\text{cylinder}} = \frac{\rho R^2}{2 \varepsilon_0 r}$

Substituting $r = 2R$: $E_{\text{cylinder}} = \frac{\rho R^2}{2 \varepsilon_0 (2R)} = \frac{\rho R}{4 \varepsilon_0}$

The direction of $\mathbf{E}_{\text{cylinder}}$ at point $P$ is radially outwards along the y-axis ($+\hat{j}$).

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Step 2: Electric Field due to the Removed Spherical Volume

The spherical cavity has a radius of $r_s = \frac{R}{2}$ with its center at the origin (on the axis of the cylinder). Point $P$ is at a distance of $r = 2R$ from the center of this sphere.

The total charge of this sphere, assuming it is uniformly filled with volume charge density $\rho$, is: $Q_{\text{sphere}} = \rho \cdot V_{\text{sphere}} = \rho \cdot \frac{4}{3} \pi \left(\frac{R}{2}\right)^3 = \frac{\pi \rho R^3}{6}$

Using Coulomb's law (or Gauss's law for a sphere) at an external point $r = 2R$: $E_{\text{sphere}} = \frac{1}{4\pi \varepsilon_0} \frac{Q_{\text{sphere}}}{r^2}$

$E_{\text{sphere}} = \frac{1}{4\pi \varepsilon_0} \frac{\frac{\pi \rho R^3}{6}}{(2R)^2} = \frac{\rho R^3}{24 \varepsilon_0 (4R^2)} = \frac{\rho R}{96 \varepsilon_0}$

The direction of $\mathbf{E}_{\text{sphere}}$ at point $P$ is also radially outwards along the y-axis ($+\hat{j}$).

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Step 3: Net Electric Field at Point $P$

By the principle of superposition, the net electric field at point $P$ is: $E_{\text{net}} = E_{\text{cylinder}} - E_{\text{sphere}}$

$E_{\text{net}} = \frac{\rho R}{4 \varepsilon_0} - \frac{\rho R}{96 \varepsilon_0}$

Taking a common denominator: $E_{\text{net}} = \frac{\rho R}{\varepsilon_0} \left( \frac{24 - 1}{96} \right) = \frac{23 \rho R}{96 \varepsilon_0}$

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Step 4: Finding the Value of $k$

Comparing the derived magnitude with the given expression $\frac{23 \rho R}{16 k \varepsilon_0}$:

$\frac{23 \rho R}{96 \varepsilon_0} = \frac{23 \rho R}{16 k \varepsilon_0}$

$16 k = 96 \implies k = 6$