An infinitely long solid cylinder of radius $R$ has a uniform volume charge density $\rho$. It has a…

An infinitely long solid cylinder of radius $R$ has a uniform volume charge density $\rho$. It has a spherical cavity of radius $R / 2$ with its centre on the axis of the cylinder, as shown in the figure. The magnitude of the electric field at the point $P$, which is at a distance $2 R$ from the axis of the cylinder, is given by the expression $\frac{23 \rho R}{16 K \varepsilon_{0}}$. The value of $k$ is

Solution

The magnitude of the electric field at the point $P$ which is at a distance $2 R$ from the axis of the cylinder $\begin{array}{l} \mathrm{E}=\mathrm{E}_{\text {total }}-\mathrm{E}_{\text {cavity }} \\ =\frac{\gamma}{2 \pi \varepsilon_{0}(2 R)}-\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{(2 R)^{2}} \\ \mathrm{Q}_{\text {sphere }}=\frac{4}{3} \pi\left(\frac{R}{2}\right)^{3} \rho=\frac{\pi R^{3} \rho}{6} \\ \begin{array}{l} \lambda_{\text {cylinder }}=\pi R^{2} \rho \\ \therefore E=\frac{\pi R^{2} \rho}{4 \pi \varepsilon_{0} R}-\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\pi R^{3} \rho / 6}{4 R^{2}} \\ =\frac{23 \rho R}{96 \varepsilon_{0}}=\frac{23 \rho R}{16 \times 6 \times \varepsilon_{0}} \\ \therefore k=6 \end{array} \end{array}$ !

Asked in: JEE Advanced 2012 (Paper 1)

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