Question:
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:
Correct Option: , 2
Solution:
flux through cube $=\frac{\mathrm{q}}{8 \epsilon_{0}}$
flux through surfaces $\mathrm{ABEH}, \mathrm{ADGH}, \mathrm{ABCD}$ will be zero
$\phi(\mathrm{EFGH})=\phi(\mathrm{DCFG})=\phi(\mathrm{EBCF})=\frac{1}{3}\left(\frac{\mathrm{q}}{8 \epsilon_{0}}\right)$
$=\frac{\mathrm{q}}{24 \epsilon_{0}}$
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