Question

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 $B$w, the flux linked with the plane surface $A$ in units of voltmeter will be:



(A) $\dfrac{q}{{{\varepsilon _0}}} - \phi $
(B) $\dfrac{1}{2}\left( {\dfrac{q}{{{\varepsilon _0}}} - \phi } \right)$
(C) $\dfrac{q}{{2{\varepsilon _0}}}$
(D) $\dfrac{\phi }{3}$

Answer 1

Hint: The flux of the wire is given by the equation of the Gauss’s law, and the net flux on the wire is given by the sum of the individual flux of the $A$, $B$ and $C$. And by using the given information in the diagram, then the flux in the point $A$ can be determined.

Useful formula
The Gauss’s law gives the relation of flux on the wire, charge on the wire and electric constant of the medium, then
$\phi = \dfrac{q}{{{\varepsilon _0}}}$
Where, $\phi $ is the electric flux of the point, $q$ is the charge in the object and ${\varepsilon _0}$ is the electric constant of the medium.

Complete step by step solution
Given that,
The charge in the hollo cylinder is, $q$
The electric flux in the hollow cylinder is, $\phi $
All the three point are in same hollow cylinder, so the net flux is equal to the sum of the individual flux of the individual points. Then,
$\phi = {\phi _A} + {\phi _B} + {\phi _C}\,.................\left( 1 \right)$
The ends of the hollow cylinder are $A$ and $C$, so the flux in these two points are same, then
${\phi _A} = {\phi _C}\,...............\left( 2 \right)$
And the net flux is equal to the flux of $B$, then
$\phi = {\phi _B}\,....................\left( 3 \right)$
Now,
The Gauss’s law gives the relation of flux on the wire, charge on the wire and electric constant of the medium, then
$\phi = \dfrac{q}{{{\varepsilon _0}}}\,................\left( 4 \right)$
Now substituting he equation (1) in the equation (4), then
${\phi _A} + {\phi _B} + {\phi _C} = \dfrac{q}{{{\varepsilon _0}}}\,...............\left( 5 \right)$
Noe substituting the equation (2) and equation (3) in the equation (5), then the above equation is written as,
${\phi _A} + \phi + {\phi _A} = \dfrac{q}{{{\varepsilon _0}}}$
By adding the terms in the above equation, then
$2{\phi _A} + \phi = \dfrac{q}{{{\varepsilon _0}}}$
By rearranging the terms in the above equation, then the above equation is written as,
$2{\phi _A} = \dfrac{q}{{{\varepsilon _0}}} - \phi $
By keeping the flux of $A$ in one side, then
${\phi _A} = \dfrac{1}{2}\left( {\dfrac{q}{{{\varepsilon _0}}} - \phi } \right)$

Hence, the option (B) is the correct answer.

Note: The net flux is equated with the flux of $B$ because in the question it is given that voltmeter shows the flux of the which is connected with the curved wire of the $B$, so the net flux is equated with the flux of $B$ and the both ends are having the same flux, so flux of $A$ and flux of $C$ are same.