I forgot this someone help

Oksi-84 [34.3K]

When a hypothesis is confirmed true by evidence from many experiments, it becomes widely accepted by scientists, and it gets a new name: 'Theory' .

6 0

3 years ago

How do I report child abuse against me online, without getting caught? I really need help S.O.S.

Veronika [31]

Incognito tab on your window gets rid of search history. Go to a website with a hotline, then call the number.

7 0

3 years ago

Read 2 more answers

A thick, 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 unc

KonstantinChe [14]

(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

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

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