Gujarati
1. Electric Charges and Fields
normal

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

A

$6$

B

$7$

C

$8$

D

$9$

(IIT-2012)

Solution

$E _1=\frac{\rho \cdot R ^2}{\varepsilon_0 \cdot 2 R } $

$E_2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\rho \cdot \frac{4}{3} \pi \cdot \frac{R^3}{8}}{(2 R)^2} $

$E_1-E_2=\frac{\rho R}{4 \varepsilon_0}-\frac{\rho . R}{\varepsilon_0 \cdot 24 \times 4} $

$=\frac{\rho R}{4 \varepsilon_0}\left[1-\frac{1}{24}\right] $

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

$\Rightarrow \quad K =6 $

Standard 12
Physics

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