A hollow cylinder has a charge of ' $q$ ' $C$ within it. If $\phi$ is the electric flux associated with the curved surface B, the flux linked with the plane surface A will be
Show Hint
Gauss's Law gives the "Total" budget of lines of force. Just subtract what is already used by one surface and split the rest between symmetric parts.
Step 1: Understanding the Concept:
Gauss's Law states that the total net electric flux ($\Phi_{\text{total}}$) passing through a closed surface is equal to the net enclosed charge divided by the permittivity of free space ($\varepsilon_0$). Step 2: Key Formula or Approach:
Gauss's Law equation:
\[ \Phi_{\text{total}} = \oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enclosed}}}{\varepsilon_0} \]
A cylinder consists of three surfaces: two flat circular plane surfaces (faces A and C) and one curved lateral surface (B).
Therefore, $\Phi_{\text{total}} = \Phi_A + \Phi_B + \Phi_C$. Step 3: Detailed Explanation:
According to Gauss's Law, for the whole cylinder enclosing charge $q$:
\[ \Phi_A + \Phi_B + \Phi_C = \frac{q}{\varepsilon_0} \]
We are given that the flux associated with the curved surface B is $\phi$, so $\Phi_B = \phi$.
Assuming the charge $q$ is symmetrically placed inside the cylinder (which is standard for such problems unless stated otherwise), the electric field distribution is symmetric.
Due to this symmetry, the flux passing through the left flat face (C) is identical to the flux passing through the right flat face (A).
So, $\Phi_C = \Phi_A$.
Substitute these into the total flux equation:
\[ \Phi_A + \phi + \Phi_A = \frac{q}{\varepsilon_0} \]
\[ 2\Phi_A + \phi = \frac{q}{\varepsilon_0} \]
Now, solve for $\Phi_A$ (the flux linked with plane surface A):
\[ 2\Phi_A = \frac{q}{\varepsilon_0} - \phi \]
\[ \Phi_A = \frac{1}{2}\left(\frac{q}{\varepsilon_0} - \phi\right) \] Step 4: Final Answer:
The flux linked with the plane surface A is $\frac{1}{2}\left(\frac{\text{q}}{\varepsilon_0} - \phi\right)$.
