IIT-JEE 2012 · Paper 1 · Physics Q16

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PhysicsInteger+4Question 16 of 60

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 $2R$ from the axis of the cylinder, is given by the expression $\dfrac{23\rho R}{16k\varepsilon_0}$ . The value of $k$ is

A long vertical cylinder of radius R along the z-axis containing a spherical cavity of radius R/2 centred on the axis at the origin; point P lies on the y-axis at distance 2R from the axis.

Your answer

Answer and worked solution

Official answer: 6

Superposition. Treat the cavity as the full cylinder plus a sphere of charge density $-\rho$ .

Cylinder. Per unit length $\lambda=\rho\pi R^2$ ; at distance $2R$ :
$E_c=\frac{\lambda}{2\pi\varepsilon_0(2R)}=\frac{\rho R}{4\varepsilon_0}.$

Cavity sphere. Charge $\rho\cdot\frac43\pi\left(\frac R2\right)^3=\frac{\rho\pi R^3}6$ ; at distance $2R$ :
$E_s=\frac{\rho\pi R^3/6}{4\pi\varepsilon_0(2R)^2}=\frac{\rho R}{96\varepsilon_0}.$

Net. $E=\dfrac{\rho R}{\varepsilon_0}\left(\dfrac14-\dfrac1{96}\right)=\dfrac{23\rho R}{96\varepsilon_0}=\dfrac{23\rho R}{16\cdot6\,\varepsilon_0}$ , so $k=6$ .

Answer: 6

Topics:Electric flux and Gauss's law Coulomb's law and superposition

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IIT-JEE 2012 · Paper 1, Physics question 16