Physics, 23.09.2019 11:20 mamaKelly2266

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?

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Answer from: Whitehouse9

(a) $E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

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:

$E(r) \cdot 4\pi r^2 = \frac{q}{\epsilon_0}$

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

q=5.00 C

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

$E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

(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

$E(r) \cdot 4\pi r^2 = \frac{q'}{\epsilon_0}$

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

We can use again Gauss theorem:

$E(r) \cdot 4\pi r^2 = \frac{q'}{\epsilon_0}$ (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

$E(r) = \frac{q}{4\pi \epsilon_0 r^2}$

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

(d) -12.3 C/m^2

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

$\sigma_a = \frac{-q}{4\pi a^2}$

where $4 \pi a^2$ 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:

$\sigma_a = \frac{-5.00 C}{4\pi (0.18 m)^2}=-12.3 C/m^2$

(e) -1.9 C/m^2

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

$\sigma_b = \frac{+q}{4\pi b^2}$

where $4 \pi b^2$ 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:

$\sigma_b = \frac{+5.00 C}{4\pi (0.46 m)^2}=-1.9 C/m^2$

Answer from: Quest

Aform of energy that is reflected or emitted from objects in the form of electrical and magnetic waves that can travel through space. there are many forms of electromagnetic energy including gamma rays, x rays, ultraviolet radiation, visible light, infrared radiation, microwaves and radio waves.

Answer from: Quest

Iwould say a is accurate

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Athick, spherical shell made of solid metal has an inner radius a = 0.18 m and an outer radius b = 0...