Question:
A disk of radius \(a/4\) having a uniformly distributed charge \(6\,\text{C}\) is placed in the x-y plane with its centre at \((-a/2, 0, 0)\).
A rod of length \(a\) carrying a uniformly distributed charge \(8\,\text{C}\) is placed on the x-axis from \(x = a/4\) to \(x = 5a/4\).
Two point charges \(-7\,\text{C}\) and \(3\,\text{C}\) are placed at \((a/4, -a/4, 0)\) and \((-3a/4, 3a/4, 0)\) respectively.
Consider a cubical surface formed by six surfaces \(x = \pm a/2\), \(y = \pm a/2\), \(z = \pm a/2\). The electric flux through this cubical surface is
A disk of radius \(a/4\) having a uniformly distributed charge \(6\,\text{C}\) is placed in the x-y plane with its centre at \((-a/2, 0, 0)\).
A rod of length \(a\) carrying a uniformly distributed charge \(8\,\text{C}\) is placed on the x-axis from \(x = a/4\) to \(x = 5a/4\).
Two point charges \(-7\,\text{C}\) and \(3\,\text{C}\) are placed at \((a/4, -a/4, 0)\) and \((-3a/4, 3a/4, 0)\) respectively.
Consider a cubical surface formed by six surfaces \(x = \pm a/2\), \(y = \pm a/2\), \(z = \pm a/2\). The electric flux through this cubical surface is
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Electric flux through a closed surface depends only on the total enclosed charge, not on its distribution.
Updated On: Mar 23, 2026
- \(-\dfrac{2C}{\varepsilon_0}\)
- \(\dfrac{2C}{\varepsilon_0}\)
- \(\dfrac{10C}{\varepsilon_0}\)
- (12C)/(varepsilon₀)
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The Correct Option is A
Solution and Explanation
Step 1: By Gauss’s law:
\[ \Phi = \frac{Q_{\text{enclosed}}}{\varepsilon_0} \]
Step 2: Disk is cut into two equal halves by the face \(x = -a/2\).
Enclosed disk charge: \(\frac{1}{2} \times 6\,\text{C} = 3\,\text{C}\).
Step 3: Only length \(a/4\) of the rod lies inside the cube.
\[ Q_{\text{rod}} = 8\,\text{C} \times \frac{1}{4} = 2\,\text{C} \]
Step 4: Of the point charges, only \(-7\,\text{C}\) lies inside the cube.
Step 5: Net enclosed charge:
\[ Q = 3\,\text{C} + 2\,\text{C} - 7\,\text{C} = -2\,\text{C} \]
\[ \Phi = \frac{Q_{\text{enclosed}}}{\varepsilon_0} \]
Step 2: Disk is cut into two equal halves by the face \(x = -a/2\).
Enclosed disk charge: \(\frac{1}{2} \times 6\,\text{C} = 3\,\text{C}\).
Step 3: Only length \(a/4\) of the rod lies inside the cube.
\[ Q_{\text{rod}} = 8\,\text{C} \times \frac{1}{4} = 2\,\text{C} \]
Step 4: Of the point charges, only \(-7\,\text{C}\) lies inside the cube.
Step 5: Net enclosed charge:
\[ Q = 3\,\text{C} + 2\,\text{C} - 7\,\text{C} = -2\,\text{C} \]
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