Question:
The distribution of some charges on two Gaussian surfaces \(A\) and \(B\) are as shown in the figure. If \(\phi_A\) and \(\phi_B\) are electric fluxes linked with the surfaces \(A\) and \(B\) respectively, then
\[
\frac{\phi_A}{\phi_B}=
\]
The distribution of some charges on two Gaussian surfaces \(A\) and \(B\) are as shown in the figure. If \(\phi_A\) and \(\phi_B\) are electric fluxes linked with the surfaces \(A\) and \(B\) respectively, then
\[
\frac{\phi_A}{\phi_B}=
\]
Show Hint
In Gauss's law problems, flux through a closed surface depends only on the net charge enclosed:
\[
\phi=\frac{q_{\text{enclosed}}}{\varepsilon_0}.
\]
Charges outside the Gaussian surface do not contribute to the net flux.
Updated On: Jun 26, 2026
- \(-\frac{1}{5}\)
- \(-3\)
- \(-\frac{3}{2}\)
- \(-\frac{3}{4}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Use Gauss's law.
According to Gauss's law, \[ \phi=\frac{q_{\text{enclosed}}}{\varepsilon_0}. \] So, electric flux depends only on the net charge enclosed by the Gaussian surface.
Step 2: Find the net charge enclosed by surface \(A\).
From the figure, surface \(A\) encloses the charges \[ +q,\quad -2q,\quad +3q,\quad -5q. \] Therefore, \[ q_A=q-2q+3q-5q. \] \[ q_A=-3q. \] Hence, \[ \phi_A=\frac{-3q}{\varepsilon_0}. \]
Step 3: Find the net charge enclosed by surface \(B\).
From the figure, surface \(B\) encloses charges whose net value is \[ q_B=4q. \] Hence, \[ \phi_B=\frac{4q}{\varepsilon_0}. \]
Step 4: Find the ratio of fluxes.
\[ \frac{\phi_A}{\phi_B} = \frac{\frac{-3q}{\varepsilon_0}}{\frac{4q}{\varepsilon_0}}. \] Canceling \(q\) and \(\varepsilon_0\), we get \[ \frac{\phi_A}{\phi_B} = -\frac{3}{4}. \]
Step 5: Final conclusion.
Therefore, \[ \boxed{-\frac{3}{4}} \] Hence, the correct option is \[ \boxed{(4)} \]
According to Gauss's law, \[ \phi=\frac{q_{\text{enclosed}}}{\varepsilon_0}. \] So, electric flux depends only on the net charge enclosed by the Gaussian surface.
Step 2: Find the net charge enclosed by surface \(A\).
From the figure, surface \(A\) encloses the charges \[ +q,\quad -2q,\quad +3q,\quad -5q. \] Therefore, \[ q_A=q-2q+3q-5q. \] \[ q_A=-3q. \] Hence, \[ \phi_A=\frac{-3q}{\varepsilon_0}. \]
Step 3: Find the net charge enclosed by surface \(B\).
From the figure, surface \(B\) encloses charges whose net value is \[ q_B=4q. \] Hence, \[ \phi_B=\frac{4q}{\varepsilon_0}. \]
Step 4: Find the ratio of fluxes.
\[ \frac{\phi_A}{\phi_B} = \frac{\frac{-3q}{\varepsilon_0}}{\frac{4q}{\varepsilon_0}}. \] Canceling \(q\) and \(\varepsilon_0\), we get \[ \frac{\phi_A}{\phi_B} = -\frac{3}{4}. \]
Step 5: Final conclusion.
Therefore, \[ \boxed{-\frac{3}{4}} \] Hence, the correct option is \[ \boxed{(4)} \]
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