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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 $A B C D ?$
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Verified Answer
The correct answer is:
$\frac{Q}{\varepsilon_{0}}$
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,
$$
\phi_{E}=\frac{Q}{\varepsilon_{0}}
$$
$$
\oint \mathbf{E} \cdot d \mathbf{S}=\frac{Q_{\mathrm{in}}}{\varepsilon_{0}}
$$
Electric flux through the five faces of the cube,
$$
\phi_{E}=\frac{Q}{\varepsilon_{0}}
$$
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