Q120 / 200 · NEET 2007 · Physics

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 $\mathrm{A}$ in units of voltmeter will be Figure

  • A
    $\frac{q}{2 \varepsilon_0}$
  • B
    $\frac{\phi}{3}$
  • C
    $\frac{q}{\varepsilon_0}-\phi$
  • Correct answer
    D
    $\frac{1}{2}\left(\frac{q}{\varepsilon_0}-\phi\right)$

Solution

Step 1: Total flux through cylinder using Gauss Law By Gauss Law, total flux through the cylinder is given by: \(\phi_{\text {total }}=\frac{\mathrm{q}}{\varepsilon_0} \ldots (1)\) Step 2: Flux through plane surface \(A\) As surfaces \(\mathrm{A} \& \mathrm{C}\) is symmetrical so \(\phi_{\mathrm{A}}=\phi_{\mathrm{C}}\) \(\phi_{\text {total }}=\phi_{\mathrm{A}}+\phi_{\mathrm{B}}+\phi_{\mathrm{C}}=\frac{\mathrm{q}}{\varepsilon_{\mathrm{o}}} \quad [\text {Using Equation (1)] }\) Given Flux through curved surface \(\phi_B=\phi\) So, \(\begin{aligned} & 2 \phi_A+\phi=\frac{q}{\varepsilon_0} \\ & \Rightarrow \phi_A=\frac{1}{2}\left(\frac{q}{\varepsilon_0}-\phi\right) \end{aligned}\)

Hence, the correct answer is option: 'D'.

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