Question

A hollow cylinder has a charge $q$ coulomb within it. If $\mathrm{\Phi }$ is the electric flux in units of voltmeter associated with the curved surface $B$, the flux linked with the plane surface in units of voltmeter will be.

Answer 1

Step 1. Given data

The charge of the hollow cylinder is $q$

Flux$=\mathrm{\Phi }$

Step 2. Formula used

According to Gaussian Law, we know that the total electric flux out in a closed surface is equal to the charge enclosed divided by the permeability

${\mathrm{\Phi }}_{}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}$

Where $\mathrm{\Phi }$ is the electric flux, $q$ is the charge enclosed and ${\mathrm{\epsilon }}_0$ is the permittivity of free space,

Step 3. Net flux through the surface

Let the flux of the surface $A,B$ and $C$ is ${\mathrm{\Phi }}_{\mathrm{A}},{\mathrm{\Phi }}_{\mathrm{B}}$ and ${\mathrm{\Phi }}_C$

Given that the electric flux in the hollow cylinder is $=\mathrm{\Phi }$

All the three points are in same hollow cylinder, so the net flux is equal to the sum of the individual flux

$\mathrm{\Phi }={\mathrm{\Phi }}_{Net}={\mathrm{\Phi }}_{\mathrm{A}}+{\mathrm{\Phi }}_{\mathrm{B}}+{\mathrm{\Phi }}_{\mathrm{C}}$

Step 4. Flux through surface $A$ ,$B$and $C$

The ends of the hollow cylinder are $A$ and $C$ so the flux in these two point are same.

$\therefore {\mathrm{\Phi }}_{\mathrm{A}}={\mathrm{\Phi }}_C$

The net flux is equal to the flux of $B$ then

Also $\mathrm{\Phi }={\mathrm{\Phi }}_{\mathrm{B}}$

Step 5. Find the flux linked with the plane surface $A$

$$\begin{gathered} {\mathrm{\Phi }}_{\mathrm{A}}+{\mathrm{\Phi }}_{\mathrm{B}}+{\mathrm{\Phi }}_{\mathrm{C}}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0} \\ \Rightarrow 2{\mathrm{\Phi }}_{\mathrm{A}}+{\mathrm{\Phi }}_{}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}\left[\because {\mathrm{\Phi }}_{\mathrm{C}}={\mathrm{\Phi }}_{\mathrm{A}}\mathrm{and}{\mathrm{\Phi }}_{\mathrm{B}}=\mathrm{\Phi }\right] \\ \Rightarrow {\mathrm{\Phi }}_{\mathrm{A}}=\frac{1}{2}\left(\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}-{\mathrm{\Phi }}_{}\right) \end{gathered}$$

Hence the flux through the surface $A$ is $\frac{1}{2}\left(\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}-{\mathrm{\Phi }}_{}\right)$