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A hollow cylinder has charge $q$ $C$ within it. If $\phi $ is the electric flux in unit of voltmeter associated with the curved surface $B$, the flux linked with the plane surface $A$ in unit of voltmeter will be
$\frac{1}{2}\left( {\frac{q}{{{\varepsilon _0}}} - \phi } \right)$
${\frac{q}{{{2\varepsilon _0}}}}$
${\frac{q}{{{\varepsilon _0}}}}$
${\frac{q}{{{\varepsilon _0}}} - \phi }$
Solution
Let electric flux linked with surfaces $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are $\phi_{\mathrm{A}}, \phi_{\mathrm{B}}$ and $\phi_{\mathrm{C}}$ respectively.
Thus $\phi_{\text {total }}=\phi_{\mathrm{A}}+\phi_{\mathrm{B}}+\phi_{\mathrm{C}}$
$ \because \phi_{A}=\phi_{C}$
and $\phi_{\text {total }}=\frac{q}{\epsilon_{0}}(\text { From Gauss's Law })$
$\frac{\mathrm{q}}{\epsilon_{0}}=2 \phi_{\mathrm{A}}+\phi_{\mathrm{B}} \quad$ But $\phi_{\mathrm{B}}=\phi$ (given)
Hence, $\frac{q}{\epsilon_{0}}=2 \phi_{A}+\phi$
or $\frac{q}{\epsilon_{0}}-\phi=2 \phi_{A}$ or $\phi_{A}=\frac{1}{2}\left(\frac{q}{\epsilon_{0}}-\phi\right)$
