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Answer 1 · 2017-04-10T20:40:42

#approx 3.7 cdot 10^10 " V m"#

The Flux is:

#Phi = int int_S mathbf E cdot d mathbf S#

And by Gauss' Law:

#int int_S mathbf E cdot d mathbf S = (sum Q)/varepsilon_o#, with: #varepsilon_o = 8.9 cdot 10^(−12) " F/m"#

Importantly, as the charge is in the centre, we can use the symmetry to say that the flux through 1 face is #Phi' = Phi/6#.

And:

#Phi' = 1/6 cdot 2/(8.9 cdot 10^(−12)) approx 3.7 cdot 10^10 " V m"#