Question 782: Electrostats & Capacitance

Question

A hollow cylinder has a charge ' $q$ ' coulomb within it. If ' $\phi$ ' is the electric flux in unit of V-m associated with the curved surface C , the flux linked with the plane surface ' $A$ ' in unit of V-m will be [ $\varepsilon_0=$ permittivity of free space]0

Accepted Answer

As per Gauss' law the net flux through closed surface is,
$$ \phi_{\mathrm{A}}+\phi_{\mathrm{B}}+\phi_{\mathrm{C}}=\frac{\mathrm{q}}{\varepsilon_0} $$
Due to symmetry, the same flux passes through plane surfaces A and B , i.e., $\phi_{\mathrm{A}}=\phi_{\mathrm{B}}$
$$ \begin{array}{lll} 	herefore & 2 \phi_{\mathrm{A}}+\phi=\frac{q}{\varepsilon_0}\,\,\,\,\,\,\,\,\,\,\,\, \ldots .\left(	ext { Given: } \phi_{\mathrm{C}}=\phiight)& \ & \phi_{\mathrm{A}}=\frac{1}{2}\left(\frac{q}{\varepsilon_0}-\phiight) \end{array} $$

Answer

$\frac{\phi}{3}$

Answer

$\left(\frac{\mathrm{q}}{\varepsilon_0}-\phi\right)$

Answer

$\quad \frac{1}{2}\left(\frac{\mathrm{q}}{\mathrm{\varepsilon}_0}-\phi\right)$

Answer

$\frac{\mathrm{q}}{2 \varepsilon_0}$

Electrostats & Capacitance

A hollow cylinder has a charge ' q ' coulomb within it. If ' \phi ' is the electric flux in unit of V-m associated with the curved surface C , the flux linked with the plane surface ' A ' in unit of V-m will be [ \varepsilon_0= permittivity of free space]0

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