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 23\rho R \over 16k\varepsilon _0 . The value of k is \underline\hspace2cm .

Electrostatics PYQ — JEE Advanced 2012 · Shift 1 · Q57 with Solution

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Correct answer: 6

Use superposition

Treat the given charge distribution as:

a complete infinite solid cylinder of radius $R$ and charge density $\rho$ ,
plus a sphere of radius $R/2$ carrying charge density $-\rho$ at the cavity location.

Since the centre of the spherical cavity lies on the axis of the cylinder, the point $P$ at distance $2R$ from the axis is also at distance $2R$ from the centre of the sphere.

So, $\vec E_{\text{net}}=\vec E_{\text{cylinder}}+\vec E_{\text{sphere with }-\rho}.$

Field due to the infinite uniformly charged cylinder at $r=2R$

For an infinite cylinder, outside the cylinder $(r>R)$ , $E_{\text{cyl}}=\frac{\lambda}{2\pi \varepsilon_0 r},$ where $\lambda=\rho \pi R^2.$

Thus,

=\frac{\rho R}{4\varepsilon_0}.$$ Direction: radially outward from the axis. --- 3. **Field due to the removed spherical cavity** The cavity is equivalent to adding a sphere of charge density $-\rho$. Its total charge is $$Q_{\text{sphere}}=-\rho\left(\frac{4}{3}\pi \left(\frac R2\right)^3\right) =-\rho\frac{\pi R^3}{6}.$$ Since point $P$ is outside the sphere ($2R > R/2$), the sphere acts like a point charge at its centre. Hence magnitude of field due to this sphere at distance $2R$ is $$E_{\text{sph}}=\frac{1}{4\pi\varepsilon_0}\cdot \frac{|Q|}{(2R)^2}.$$ Substitute: $$E_{\text{sph}}=\frac{1}{4\pi\varepsilon_0}\cdot \frac{\pi R^3\rho/6}{4R^2} =\frac{\rho R}{96\varepsilon_0}.$$ Because the sphere has **negative** charge, its field at $P$ is directed **towards** the sphere centre, i.e. opposite to the cylinder field. --- 4. **Net electric field** Therefore, $$E_{\text{net}}=E_{\text{cyl}}-E_{\text{sph}} =\frac{\rho R}{4\varepsilon_0}-\frac{\rho R}{96\varepsilon_0}.$$ Taking LCM: $$E_{\text{net}}=\frac{24\rho R-\rho R}{96\varepsilon_0} =\frac{23\rho R}{96\varepsilon_0}.$$ Given, $$E=\frac{23\rho R}{16k\varepsilon_0}.$$ Equate: $$\frac{23\rho R}{16k\varepsilon_0}=\frac{23\rho R}{96\varepsilon_0}.$$ So, $$16k=96 \implies k=6.$$ --- 5. **Final answer** $$\boxed{k=6}$$

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View written solutionFree

Correct answer: 6

Use superposition

Treat the given charge distribution as:

a complete infinite solid cylinder of radius $R$ and charge density $\rho$ ,
plus a sphere of radius $R/2$ carrying charge density $-\rho$ at the cavity location.

Since the centre of the spherical cavity lies on the axis of the cylinder, the point $P$ at distance $2R$ from the axis is also at distance $2R$ from the centre of the sphere.

So, $\vec E_{\text{net}}=\vec E_{\text{cylinder}}+\vec E_{\text{sphere with }-\rho}.$

Field due to the infinite uniformly charged cylinder at $r=2R$

For an infinite cylinder, outside the cylinder $(r>R)$ , $E_{\text{cyl}}=\frac{\lambda}{2\pi \varepsilon_0 r},$ where $\lambda=\rho \pi R^2.$

Thus,

=\frac{\rho R}{4\varepsilon_0}.$$ Direction: radially outward from the axis. --- 3. **Field due to the removed spherical cavity** The cavity is equivalent to adding a sphere of charge density $-\rho$. Its total charge is $$Q_{\text{sphere}}=-\rho\left(\frac{4}{3}\pi \left(\frac R2\right)^3\right) =-\rho\frac{\pi R^3}{6}.$$ Since point $P$ is outside the sphere ($2R > R/2$), the sphere acts like a point charge at its centre. Hence magnitude of field due to this sphere at distance $2R$ is $$E_{\text{sph}}=\frac{1}{4\pi\varepsilon_0}\cdot \frac{|Q|}{(2R)^2}.$$ Substitute: $$E_{\text{sph}}=\frac{1}{4\pi\varepsilon_0}\cdot \frac{\pi R^3\rho/6}{4R^2} =\frac{\rho R}{96\varepsilon_0}.$$ Because the sphere has **negative** charge, its field at $P$ is directed **towards** the sphere centre, i.e. opposite to the cylinder field. --- 4. **Net electric field** Therefore, $$E_{\text{net}}=E_{\text{cyl}}-E_{\text{sph}} =\frac{\rho R}{4\varepsilon_0}-\frac{\rho R}{96\varepsilon_0}.$$ Taking LCM: $$E_{\text{net}}=\frac{24\rho R-\rho R}{96\varepsilon_0} =\frac{23\rho R}{96\varepsilon_0}.$$ Given, $$E=\frac{23\rho R}{16k\varepsilon_0}.$$ Equate: $$\frac{23\rho R}{16k\varepsilon_0}=\frac{23\rho R}{96\varepsilon_0}.$$ So, $$16k=96 \implies k=6.$$ --- 5. **Final answer** $$\boxed{k=6}$$

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