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 :

• A. $\frac{ q }{4 \varepsilon_{0}}$
• B. $\frac{ q }{24 \varepsilon_{0}}$
• C. $\frac{ q }{48 \varepsilon_{0}}$
• D. $\frac{ q }{8 \varepsilon_{0}}$

Answer 1

Answer: (B)

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