Q.
A disc of radius $\frac{a}{4}\ $having a uniformly distributed charge 6C 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 C is placed on the $x$-axis form $x = \frac{a}{4}\ to\ x = \frac{5a}{4}.$ Two point charges -7C and 3C are placed at $\left( \frac{a}{4},\frac{- a}{4},0 \right)and\ \left( \frac{- 3a}{4},\frac{3a}{4},0 \right)$ 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
A) $\frac{- 2C}{\varepsilon_{0}}$
B) $\frac{2C}{\varepsilon_{0}}$
C) $\frac{10C}{\varepsilon_{0}}$
D) $\frac{12C}{\varepsilon_{0}}$
Explanation
(a) Total enclosed charge as already shown in
$q_{net} = \frac{6C}{2} + \frac{8C}{4} - 7C$
$\ \ \ \ \ \ \ \ \ = - 2C$
From Gauss-theorem, net flux,
$\phi_{net} = \frac{q_{net}}{\varepsilon_{0}} = \frac{- 2C}{\varepsilon_{0}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.