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JEE Advanced · Physics · 17. Electrostatics

A disk of radius \(\frac{a}{4}\) having a uniformly distributed charge \(6 \mathrm{C}\) is placed in the \(x-y\) plane with its centre at \(\left(\frac{-a}{2}, 0,0\right)\). A rod of length a carrying a uniformly distributed charge \(8 \mathrm{C}\) is placed on the \(x\)-axis from \(x=\frac{a}{4}\) to \(x=\frac{5 a}{4}\). Two point charges \(-7 \mathrm{C}\) and \(3 \mathrm{C}\) are placed at \(\left(\frac{a}{4}, \frac{-a}{4}, 0\right)\) and \(\left(\frac{-3 a}{4}, \frac{3 a}{4}, 0\right)\), respectively. Consider a cubical surface formed by six surfaces \(x=\pm \frac{a}{2}, y=\pm \frac{a}{2}, z=\pm \frac{a}{2}\).
The electric flux through this cubical surface is

  1. A \(\frac{-2 C}{\varepsilon_0}\)
  2. B \(\frac{2 C}{\varepsilon_0}\)
  3. C \(\frac{10 C}{\varepsilon_0}\)
  4. D \(\frac{12 C}{\varepsilon_0}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{-2 C}{\varepsilon_0}\)

Step-by-step Solution

Detailed explanation

Total enclosed charge as already shown is
\(
q_{\text {net }}=\frac{6 C}{2}+\frac{8 C}{4}-7 C=-2 C
\)
From Gauss-theorem, net flux,
\(
\varphi_{\text {net }}=\frac{q_{\text {net }}}{\varepsilon_0}=\frac{-2 C}{\varepsilon_0}
\)
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