Rankers Physics

Gauss's Law: Practice Problem & Solution

A hollow cylinder has a charge $q$ coulomb within it. If $\phi$ is the electric flux in units of voltmeter associated with the curved surface B, the flux linked with the plane surface A in units of voltmeter will be: (2007)
$\frac{1}{2} (\frac{q}{\epsilon_0} - \phi)$
$\frac{q}{2\epsilon_0}$
$\frac{\phi}{3}$
$\frac{q}{\epsilon_0} - \phi$

Solution Explained:

To solve this problem, we apply the core principles of Gauss's Law. Understanding the underlying formula is key to arriving at the correct answer below:

Total electric flux through the entire hollow cylinder is $\Phi_{total} = \frac{q}{\epsilon_0}$. This total flux is the sum of fluxes through the two plane surfaces (A and C) and the curved surface (B). By symmetry, $\Phi_A = \Phi_C$. Thus, $2\Phi_A + \phi = \frac{q}{\epsilon_0}$, which gives $\Phi_A = \frac{1}{2}(\frac{q}{\epsilon_0} - \phi)$.

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