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.

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Solution

## Step 1. Given dataThe charge of the hollow cylinder is $q$Flux$=\mathrm{\Phi }$Step 2. Formula usedAccording 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 surfaceLet 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 $B$ thenAlso $\mathrm{\Phi }={\mathrm{\Phi }}_{\mathrm{B}}$Step 5. Find the flux linked with the plane surface $A$${\mathrm{\Phi }}_{\mathrm{A}}+{\mathrm{\Phi }}_{\mathrm{B}}+{\mathrm{\Phi }}_{\mathrm{C}}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_{0}}\phantom{\rule{0ex}{0ex}}⇒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]\phantom{\rule{0ex}{0ex}}⇒{\mathrm{\Phi }}_{\mathrm{A}}=\frac{1}{2}\left(\frac{\mathrm{q}}{{\mathrm{\epsilon }}_{0}}-{\mathrm{\Phi }}_{}\right)$Hence the flux through the surface $A$ is $\frac{1}{2}\left(\frac{\mathrm{q}}{{\mathrm{\epsilon }}_{0}}-{\mathrm{\Phi }}_{}\right)$

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