Question

A hollow cylinder has a charge $q$ within it. If $\varphi$ is the electric flux associated with curved surface $B.$ The flux linked with the place surface $A$ will be

• A. $\frac{q}{{\epsilon }_0}-\varphi$
• B. $\frac{1}{2}\left[\frac{q}{{\epsilon }_0}-\varphi \right]$
• C. $\frac{q}{2{\epsilon }_0}$
• D. $\frac{\varphi }{3}$

Answer 1

The correct option is B $\frac{1}{2}\left[\frac{q}{{\epsilon }_0}-\varphi \right]$

According to gauss law,

$\varphi =\frac{Q_{inc}}{{\epsilon }_0}⟶\left(1\right)$

& from the definition of $\varphi$

$\varphi =E.da⟶\left(2\right)$

For cylindrical surface $E.da=2E.\{Ar.A\}⟶\left(3\right)$

$+E.\{Ar.B\}$

where $\left\{ArA\right\}$ $=$ Area of surface $A$.

Also, $ATQ$, $E.\left\{ArB\right\}={\varphi }^1⟶\left(4\right)$

$\therefore$ From equ (1), (2), (3) & (4)

$2E.\{Ar.A\}+{\varphi }^1=\frac{q}{{\epsilon }_0}$

or $E.\{Ar.A\}=\{\frac{q}{{\epsilon }_0}-\varphi \}\frac{1}{2}$

$\therefore$ flux through surface area, $A$, is given by,

$E.\left\{ArA\right\}=\frac{1}{2}\{\frac{q}{{\epsilon }_0}-\varphi \}$

$\therefore$ Option (B) is the correct answer.