Athick, spherical shell made of solid metal has an inner radius a = 0.18 m and an outer radius b = 0.33 m, and is initially uncharged. a point charge q = 5.00 c is placed at the center of the shell. what is the electric field strength in the region r < a? express your answer in terms of 1/r2. tries 0/8 what is the electric field strength in the region a < r < b? express your answer in terms of 1/r2. tries 0/8 what is the electric field strength in the region b < r? express your answer in terms of 1/r2. tries 0/8 what is the induced charge density at r = a? (in c/m^2) tries 0/8 what is the induced charge density (in c/m2) at r = b? (in c/m^2)

A)

r < a

We can find the magnitude of the electric field by using Gauss theorem. Taking a Gaussian spherical surface of radius r centered in the centre of the sphere, the electric flux through the surface of the sphere is equal to the ratio between the charge contained in the sphere and the vacuum permittivity:

For r < a, the charge contained in the gaussian sphere is the point charge:

So the electric field in this region is

B) E = 0

a < r < b

The region a < r < b is the region between the inner and the outer surface of the shell. We have to keep in mind that the presence of the single point charge +q = 5.00 C at the center of the sphere induces an opposite charge -q on the inner surface (r=a), and a charge of +q at the outer surface (r=b).

Using again Gauss theorem

this time we have that the gaussian sphere contains both the single point charge +q and the negative charge -q induced at r=a, so the net charge contained in the sphere is

q' = +q - q = 0

And so, the electric field in this region is zero.

C)

r > b

Here we are outside of the sphere. Using Gauss theorem again

this time we have that the gaussian sphere contains the single point charge +q, the negative charge -q induced at r=a, and the positive charge +q induced at r=b, so the net charge contained in the sphere is

q' = +q - q +q = q

And so the electric field is identical to the one inside the sphere:

D) -12.29 C/m^2

We said that the charge induced at the inner surface r=a is

-q = -5.00 C

The induced charge density is

where A is the area of the inner surface of radius r = a = 0.18 m, so it is

So the induced charge density is

E) +3.65 C/m^2

We said that the charge induced at the outer surface r=b is

+q = +5.00 C

The induced charge density is

where A is the area of the outer surface of radius r = b = 0.33 m, so it is

So the induced charge density is

i think if it were me i whould divide 300 dide 15

step-by-step explanation:

ice to water 1 water to vapour 2 latent heat