JEE Advanced 2012Paper 1
- Exam
- JEE Advanced
- Year
- 2012
- Paper
- Paper 1
- Subject
- Physics
- Chapter
- Electrostatics
JEE Advanced 2012 · Paper 1 · Physics · Electrostatics
An infinitely long solid cylinder of radius R has a uniform volume...
An infinitely long solid cylinder of radius has a uniform volume charge density . It has a spherical cavity of radius with its centre on the axis of the cylinder, as shown in the figure. The magnitude of the electric field at the point , which is at a distance from the axis of the cylinder, is given by the expression . The value of is
Correct answer
0
Step-by-step solution
The magnitude of the electric field at the point which is at a distance 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}$