MHT CET · Physics · Electrostatics
A hollow cylinder has a charge \(q\) coulomb within it. If \(\phi\) is the electric flux associated with the surface \(B\), the flux linked with the plane surface A will be
- A \(\frac{\phi}{2}\)
- B \(\frac{\phi}{\epsilon_0}-\phi\)
- C \(\frac{1}{2}\left(\frac{\mathrm{q}}{\epsilon_0}-\phi\right)\)
- D \(\frac{q}{2 \epsilon_0}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{2}\left(\frac{\mathrm{q}}{\epsilon_0}-\phi\right)\)
Step-by-step Solution
Detailed explanation
The correct option is (C).
Concept: Gauss Law: Net flux through an enclosing surface is proportional to the charge inside, or \(\phi_{\mathrm{T}}=\frac{\mathrm{q}}{\varepsilon_0}\), where \(\mathrm{q}\) is the charge inside, \(\varepsilon_0\) the free permitivity of the space and \(\phi_{\mathrm{T}}\) is the total flux
The flux associated with the lateral surface B is \(\phi\). Due to symmetry of the system, the flux associated with surface A should be the same as surface \(\mathrm{C}\). Let the flux associated with the surface A be \(\phi_0\).
Therefore, the total flux is \(\phi_0+\phi+\phi_0=\frac{q}{\varepsilon_0}\)
So, the flux associated with the surface \(A\) is \(\phi_0=\frac{1}{2}\left(\frac{q}{\epsilon_0}-\phi\right)\)
Concept: Gauss Law: Net flux through an enclosing surface is proportional to the charge inside, or \(\phi_{\mathrm{T}}=\frac{\mathrm{q}}{\varepsilon_0}\), where \(\mathrm{q}\) is the charge inside, \(\varepsilon_0\) the free permitivity of the space and \(\phi_{\mathrm{T}}\) is the total flux
The flux associated with the lateral surface B is \(\phi\). Due to symmetry of the system, the flux associated with surface A should be the same as surface \(\mathrm{C}\). Let the flux associated with the surface A be \(\phi_0\).
Therefore, the total flux is \(\phi_0+\phi+\phi_0=\frac{q}{\varepsilon_0}\)
So, the flux associated with the surface \(A\) is \(\phi_0=\frac{1}{2}\left(\frac{q}{\epsilon_0}-\phi\right)\)
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