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A hollow cylinder has a charge $q$ coulomb within it. If $\phi$ is the electric flux in units of voltmeter associated with the curved surface B, the flux linked with the plane surface $\mathrm{A}$ in units of voltmeter will be
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Verified Answer
The correct answer is:
$\frac{1}{2}\left(\frac{q}{\varepsilon_0}-\phi\right)$
Step 1: Total flux through cylinder using Gauss Law
By Gauss Law, total flux through the cylinder is given by:
\(\phi_{\text {total }}=\frac{\mathrm{q}}{\varepsilon_0} \ldots (1)\)
Step 2: Flux through plane surface \(A\)
As surfaces \(\mathrm{A} \& \mathrm{C}\) is symmetrical so \(\phi_{\mathrm{A}}=\phi_{\mathrm{C}}\)
\(\phi_{\text {total }}=\phi_{\mathrm{A}}+\phi_{\mathrm{B}}+\phi_{\mathrm{C}}=\frac{\mathrm{q}}{\varepsilon_{\mathrm{o}}} \quad [\text {Using Equation (1)] }\)
Given Flux through curved surface \(\phi_B=\phi\)
So,
\(\begin{aligned}
& 2 \phi_A+\phi=\frac{q}{\varepsilon_0} \\
& \Rightarrow \phi_A=\frac{1}{2}\left(\frac{q}{\varepsilon_0}-\phi\right)
\end{aligned}\)
By Gauss Law, total flux through the cylinder is given by:
\(\phi_{\text {total }}=\frac{\mathrm{q}}{\varepsilon_0} \ldots (1)\)
Step 2: Flux through plane surface \(A\)
As surfaces \(\mathrm{A} \& \mathrm{C}\) is symmetrical so \(\phi_{\mathrm{A}}=\phi_{\mathrm{C}}\)
\(\phi_{\text {total }}=\phi_{\mathrm{A}}+\phi_{\mathrm{B}}+\phi_{\mathrm{C}}=\frac{\mathrm{q}}{\varepsilon_{\mathrm{o}}} \quad [\text {Using Equation (1)] }\)
Given Flux through curved surface \(\phi_B=\phi\)
So,
\(\begin{aligned}
& 2 \phi_A+\phi=\frac{q}{\varepsilon_0} \\
& \Rightarrow \phi_A=\frac{1}{2}\left(\frac{q}{\varepsilon_0}-\phi\right)
\end{aligned}\)
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