Question 296 of 311

A charge $q$ is placed at one corner of a cube as shown in figure. The flux of electrostatic field $\vec{E}$ through the shaded area is:

A$\dfrac{q}{8{\epsilon }_{0}}$B$\dfrac{q}{24{\epsilon }_{0}}$C$\dfrac{q}{48{\epsilon }_{0}}$D$\dfrac{q}{4{\epsilon }_{0}}$

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Answer: (B) $\dfrac{q}{24{\epsilon }_{0}}$

flux through cube $=\dfrac{q}{8{ϵ}_{0}}$
flux through surfaces $ABEH,ADGH,ABCD$ will be zero
$\phi \left(EFGH\right)=\phi \left(DCFG\right)=\phi \left(EBCF\right)=\dfrac{1}{3}\left(\dfrac{q}{8{ϵ}_{0}}\right)$
$=\dfrac{q}{24{ϵ}_{0}}$

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