Question

Two concentric conducting spherical shells produce a radially outward electric field of magnitude 49,000 N/C at a point 4.10 m from the center of the shells. The outer surface of the larger shell has a radius of 3.75 m. If the inner shell contains an excess charge of -5.30 μC, find the amount of charge on the outer surface of the larger shell. (k = 1/(4πε0) = 8.99 × 109 N · m2/C2)

Answer 1

Answer: the electric charge in the outher layer is = 97.12 μC

Step-by-step 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