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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
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The correct answer is:
$\frac{1}{2}\left(\frac{\mathrm{q}}{\epsilon_0}-\phi\right)$
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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