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)

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

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2

no

sodium is solid at rt