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Question

Let ${\phi _A},{\phi _B}$ and ${\phi _C}$ are the electric flux linked with $A,B$ and $C.$

According to gauss theorem,

${\phi _A} + {\phi _B} +

Vedclass · Accepted Answer

$\frac{1}{2}$($\frac{q}{{{\varepsilon _0}}}$ -$\phi$)

Vedclass · Answer

Let ${\phi _A},{\phi _B}$ and ${\phi _C}$ are the electric flux linked with $A,B$ and $C.$

According to gauss theorem,

${\phi _A} + {\phi _B} + {\phi _C} = \frac{q}{{{\varepsilon _0}}}$

$\sin ce\,{\phi _A} = {\phi _C},$

$	herefore \,2{\phi _A} + {\phi _B} = \frac{q}{{{\varepsilon _0}}}\,\,\,or\,\,2{\phi _A} = \frac{q}{{{\varepsilon _0}}} - {\phi _B}$

or,  $2{\phi _A} = \frac{q}{{{\varepsilon _0}}} - \phi $

(Given ${\phi _B} = \phi $).

$	herefore {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\phi _A} = \frac{1}{2}\left( {\frac{q}{{{\varepsilon _0}}} - \phi } ight).$

A hollow cylinder has a charge $q$ coulomb within it. If $\phi$ is the electric flux in units of $volt-meter$ associated with the curved surface $B,$ the flux linked with the plane surface $A$ in units of $V-m$ will be

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