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. \(-\dfrac{2C}{\varepsilon_0}\)
• B. \(\dfrac{2C}{\varepsilon_0}\)
• C. \(\dfrac{10C}{\varepsilon_0}\)
• D. (12C)/(varepsilon₀)

Hint:

Electric flux through a closed surface depends only on the total enclosed charge, not on its distribution.

Answer 1

The Correct Option is A

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:
\[ Q = 3\,\text{C} + 2\,\text{C} - 7\,\text{C} = -2\,\text{C} \]