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MHT CET · Physics · Electrostatics

A hollow cylinder has charge ' \(q\) ' \(C\) within it. If ' \(\phi\) ' is the electric flux associated with the curved surface B , the flux linked with the plane surface A will be

  1. A \(\frac{1}{2}\left(\frac{q}{\varepsilon_0}-\phi\right)\)
  2. B \(\frac{q}{2 \varepsilon_0}\)
  3. C \(\frac{\phi}{3}\)
  4. D \(\frac{q}{\varepsilon_0}-\phi\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{1}{2}\left(\frac{q}{\varepsilon_0}-\phi\right)\)

Step-by-step Solution

Detailed explanation


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 C, i.e., \(\phi_{\mathrm{A}}=\phi_{\mathrm{C}}\)
\(\begin{array}{ll}
\therefore \quad & 2 \phi_A+\phi=\frac{q}{\varepsilon_0} \\
& \phi_A=\frac{1}{2}\left(\frac{q}{\varepsilon_0}-\phi\right)
\end{array}\)
\(\ldots\left(\text { Given: } \phi_B=\phi\right)\)