A) \[\frac{2q}{{{\varepsilon }_{0}}}\]
B) \[\frac{q}{8{{\varepsilon }_{0}}}\]
C) \[\frac{q}{{{\varepsilon }_{0}}}\]
D) \[\frac{q}{2{{\varepsilon }_{0}}}6{{a}^{2}}\]
Correct Answer: B
Solution :
Charge enclosed = q / 8 Therefore, flux \[\phi =\frac{{{q}_{enclosed}}}{{{\varepsilon }_{0}}}\] \[\phi =\frac{q}{8{{\varepsilon }_{0}}}\]
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