Answer: the electric charge in the outher layer is = 97.12 μC
Explanation:
The equation for electric field E caused by an electric charge is:
E= K*Q/d^(2) (1)
Where K = 8.99*10^(9) Nm^(2)/C^(2) K = 1/4*π*ε₀
Q is the electric charge and d is distance between the electric charge and the point where the field is requiered
In this case by simetry the electric field produced for 2 concentric shells could be calculated as the electric charge where at the center of the concentric shell.
the shells produce and outward field and the inner shell have a charge of -5,30 μC; this mean that the outside shell must have a positive ( an greater charge) in order to get an outward field
So:
E = 49000 N/C
d = 4.10 m then d^(2) is (4.10)^(2) = 16.81 m^(2)
q electric charge in the inner shell -5.5 μC o -5,5* 10^(-6)
Then using a gaussian surface (an sphere with the same center as the shells we are able to apply equation (1) and get the net charge of the shells
E = K* Q(n) / r^(2) Q (n) = E* r^(2) /K
Q (n) = (49000* 16.81 Nm^(2)/C )/8.99*10^(9) Nm^(2)/C^(2)
Q(n) = 91622* 10^(-9)C or 91,62 10^(-6)C or 91.62 μC
Then the charge in the outside shell is 91,62 + 5,5 μC = 97.12 μC
