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Q.
In figure charge is located at one of the.edge of the cube, then electric flux through cube due to +Q charge is
Bihar CECEBihar CECE 2009Electric Charges and Fields
Solution:
assuming 3 more cubes to make it symmetric
total flux through the 4 cube
using geuss law
$\phi_{\text {total }}=\frac{\phi_{\text {mide }}}{\epsilon_{0}}=\frac{+ q}{\epsilon_{0}} $
$\phi_{\text {each }}=\frac{\phi_{\text {total }}}{4}=\frac{+q}{4 \epsilon_{0}}$