357962
An infinitely long solid cylinder of radius \(R\) has a uniform charge density \(\rho \). It has a spherical cavity of radius \(\frac{R}{2}\) with its centre on the axis of the cylinder (as shown in figure). The magnitude of the electric field at point P, which is at a distance \(2R\) from the axis of the cylinder, is given by \(\frac{{23\rho R}}{{6K{\varepsilon _0}}}.\) Find the value of \(K\)
1 16
2 22
3 36
4 19
Explanation:
Electric field at \(P\) due to cylinder, \({E_1} = \frac{\lambda }{{2\pi {\varepsilon _0}(2R)}} = \frac{{\rho \pi {R^2}}}{{4\pi {\varepsilon _0}R}} = \frac{{\rho R}}{{4{\varepsilon _0}}}\) \(\,\,\,\,\,\,\left( {\because \lambda = \frac{Q}{h} = \frac{{\rho \pi {r^2}h}}{h} = \rho \pi {r^2}} \right)\) Electric field at \({P}\) due to spherical cavity, \({E}_{2}=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{\rho \cdot \dfrac{4}{3} \pi \dfrac{{R}^{3}}{8}}{(2 {R})^{2}}=\dfrac{\rho {R}}{(24 \times 4) \varepsilon_{0}}\) \({E_{net{\rm{ }}}} = {E_1} - {E_2}\) \( = \frac{{\rho R}}{{4{\varepsilon _0}}} - \frac{{\rho R}}{{96{\varepsilon _0}}}\) \( = \frac{{\rho R}}{{4{\varepsilon _0}}}\left( {1 - \frac{1}{{24}}} \right)\) \( = \frac{{23}}{{96}}\frac{{\rho R}}{{{\varepsilon _0}}} = \frac{{23}}{{6 \times 16}}\frac{{\rho R}}{{{\varepsilon _0}}}\) \(\therefore \quad K=16\)
