Q161 / 180 · COMEDK 2017 · Physics

The electric flux for Gaussian surface $A$ that enclose the charged particles in free space is (Given, $q_{1}=-14 \mathrm{nC}, q_{2}=78.85 \mathrm{nC}, q_{3}=-56 \mathrm{nC}$ )
Figure

  • A
    $10^{3} \mathrm{~N}-\mathrm{m}^{2} \mathrm{C}^{-1}$
  • Correct answer
    B
    $10^{3} \mathrm{CN}^{-1} \mathrm{~m}^{-2}$
  • C
    $6.32 \times 10^{3} \mathrm{~N}-\mathrm{m}^{2} \mathrm{C}^{-1}$
  • D
    $6.32 \times 10^{3} \mathrm{CN}^{-1} \mathrm{~m}^{-2}$

Solution

Given, $\begin{aligned} &q_{1}=-14 \mathrm{nC}=-14 \times 10^{-9} \mathrm{C} \\ &q_{2}=78.85 \mathrm{nC}=78.85 \times 10^{-9} \mathrm{C} \\ &q_{3}=-56 \mathrm{nC}=-56 \times 10^{-9} \mathrm{C} \end{aligned}$ According to Gauss's law Net electric flux passing through closed surface $A$, $\begin{aligned} \phi_{A} &=\frac{1}{\varepsilon_{0}}(\text { total charged enclosed by closed surface } A) \\ &=\frac{1}{\varepsilon_{0}} \cdot\left(q_{1}+q_{2}+q_{3}\right) \end{aligned}$ $\begin{aligned} &=\frac{-14 \times 10^{-9}+78.85 \times 10^{-9}-56 \times 10^{-9}}{8.85 \times 10^{-12}} \\ &=\frac{8.85}{8.85} \times 10^{-9+12}=10^{3} \mathrm{CN}^{-1} \mathrm{~m}^{-2} \end{aligned}$

Hence, the correct answer is option: 'B'.

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