Q139 / 155 · WBJEE 2018 · Physics

Consider a region in free space bounded by the surfaces of an imaginary cube having sides of length $a$ as shown in the figure. A charge $+Q$ is placed at the centre $O$ of the cube. $P$ is such a point out side the cube that the line $O P$ perpendicularly intersects the surface $A B C D$ at $R$ and also $O R=R P=a / 2$. A charge $+Q$ is placed at point $P$ also. What is the total electric flux through the five faces of the cube other than $A B C D ?$ Figure

  • Correct answer
    A
    $\frac{Q}{\varepsilon_{0}}$
  • B
    $\frac{5 Q}{6 \varepsilon_{0}}$
  • C
    $\frac{10 Q}{6 \varepsilon_{0}}$
  • D
    zero

Solution

For surface $A B C D$ clectric flux is zero. Because at surface $A B C D$ net electric field is zero. Using Gauss's law, $$ \oint \mathbf{E} \cdot d \mathbf{S}=\frac{Q_{\mathrm{in}}}{\varepsilon_{0}} $$ Electric flux through the five faces of the cube, $$ \phi_{E}=\frac{Q}{\varepsilon_{0}} $$

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

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