Athick, spherical shell made of solid metal has an inner radius a = 0.18 m and an outer radius b = 0.46 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?

(a)

We can solve the different part of the problem by using Gauss theorem.

Considering a Gaussian spherical surface with radius r<a (inside the shell), we can write:

where q is the charge contained in the spherical surface, so

Solving for E(r), we find the expression of the field for r<a:

(b) 0

The electric field strength in the region a < r < b is zero. This is due to the fact that the charge +q placed at the center of the shell induces an opposite charge -q on the inner surface of the shell (r=a), while the outer surface of the shell (r=b) will acquire a net charge of +q.

So, if we use Gauss theorem for the region  a < r < b, we get

however, the charge q' contained in the Gaussian sphere of radius r is now the sum of the charge at the centre (+q) and the charge induced on the inner surface of the shell (-q), so

q' = + q - q = 0

And so we find

E(r) = 0

(c)

We can use again Gauss theorem:

(1)

where this time r > b (outside the shell), so the gaussian surface this time contained:

- the charge +q at the centre

- the inner surface, with a charge of -q

- the outer surface, with a charge of +q

So the net charge is

q' = +q -q +q = +q

And so solving (1) we find

which is identical to the expression of the field inside the shell.

(d)

We said that at r = a, a charge of -q is induced. The induced charge density will be

where is the area of the inner surface of the shell. Substituting

q = 5.00 C

a = 0.18 m

We find the induced charge density:

(e)

We said that at r = b, a charge of +q is induced. The induced charge density will be

where is the area of the outer surface of the shell. Substituting

q = 5.00 C

b = 0.46 m

We find the induced charge density:

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000