Question
A hollow cylinder has a charge $$q$$ coulomb within it. If $$\phi $$ is the electric flux in unit of voltmeter associated with the curved surface $$B,$$ the flux linked with the plane surface $$A$$ in unit of voltmeter will be
A hollow cylinder has a charge $$q$$ coulomb within it. If $$\phi $$ is the electric flux in unit of voltmeter associated with the curved surface $$B,$$ the flux linked with the plane surface $$A$$ in unit of voltmeter will be
A.
$$\frac{1}{2}\left( {\frac{q}{{{\varepsilon _0}}} - \phi } \right)$$
B.
$$\frac{q}{{2{\varepsilon _0}}}$$
C.
$$\frac{\phi }{3}$$
D.
$$\frac{q}{{{\varepsilon _0}}} - \phi $$
Answer :
$$\frac{1}{2}\left( {\frac{q}{{{\varepsilon _0}}} - \phi } \right)$$
Solution :
Gauss’s law states that the net electric flux through any closed surface is equal to the net charge inside the surface divided by $${{\varepsilon _0}}.$$
i.e. $${\phi _{{\text{total }}}} = \frac{{{q_{{\text{inside }}}}}}{{{\varepsilon _0}}}$$
Let electric flux linked with surfaces $$A,B$$ and $$C$$ are $${\phi _A},{\phi _B}$$ and $${\phi _C}$$ respectively, i.e.
$$\eqalign{ & {\phi _{{\text{total }}}} = {\phi _A} + {\phi _B} + {\phi _C} \cr & {\text{Since,}}\,\,{\phi _C} = {\phi _A} \cr & \therefore 2{\phi _A} + {\phi _B} = {\phi _{{\text{total}}}} = \frac{q}{{{\varepsilon _0}}} \cr & {\text{or}}\,\,{\phi _A} = \frac{1}{2}\left( {\frac{q}{{{\varepsilon _0}}} - {\phi _B}} \right) \cr & {\text{But}}\,\,{\phi _B} = \phi \,\,\left( {{\text{given}}} \right) \cr & {\text{Hence,}}\,\,{\phi _A} = \frac{1}{2}\left( {\frac{q}{{{\varepsilon _0}}} - \phi } \right) \cr} $$
Gauss’s law states that the net electric flux through any closed surface is equal to the net charge inside the surface divided by $${{\varepsilon _0}}.$$
i.e. $${\phi _{{\text{total }}}} = \frac{{{q_{{\text{inside }}}}}}{{{\varepsilon _0}}}$$
Let electric flux linked with surfaces $$A,B$$ and $$C$$ are $${\phi _A},{\phi _B}$$ and $${\phi _C}$$ respectively, i.e.
$$\eqalign{ & {\phi _{{\text{total }}}} = {\phi _A} + {\phi _B} + {\phi _C} \cr & {\text{Since,}}\,\,{\phi _C} = {\phi _A} \cr & \therefore 2{\phi _A} + {\phi _B} = {\phi _{{\text{total}}}} = \frac{q}{{{\varepsilon _0}}} \cr & {\text{or}}\,\,{\phi _A} = \frac{1}{2}\left( {\frac{q}{{{\varepsilon _0}}} - {\phi _B}} \right) \cr & {\text{But}}\,\,{\phi _B} = \phi \,\,\left( {{\text{given}}} \right) \cr & {\text{Hence,}}\,\,{\phi _A} = \frac{1}{2}\left( {\frac{q}{{{\varepsilon _0}}} - \phi } \right) \cr} $$
