Question

The electric flux through each of the six sides of a rectangular box are as follows: Φ1 = 131.6 N⋅m2/C , Φ2 = 251.2 N⋅m2/C , Φ3 = -348.5 N⋅m2/C , Φ4 = 166.8 N⋅m2/C , Φ5 = -108.6 N⋅m2/C , Φ6 = 461.6 N⋅m2/C . How much charge is in this box?

Answer 1

From Gauss law

the electric flux through a closed surface = (1/epsilon0)Q_in

here the given fluxes through the faces of the rectangular box is

Φ1 = 131.6 N⋅m2/C , Φ2 = 251.2 N⋅m2/C , Φ3 = -348.5 N⋅m2/C , Φ4 = 166.8 N⋅m2/C , Φ5 = -108.6 N⋅m2/C , Φ6 = 461.6 N⋅m2/C

so the net flux is  

=1+​2+3+​45+​6

phi = 131.6+251.2-348.5+166.8-108.6+461.6 = 554.1 N⋅m2/C

from GAuss law

554.1 = (Q_in)/8.854*10^-12

Q_in = 4.9060014*10^-9 C

Q_in = 4.9060 nC

charge is in this box = 4.9060 nC