A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 4.05 m/s. If the roof is pitched at
1 answer:
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(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:
Answer: (a) and (b) => check attached file.
(c). Picture (a) and (b) will both remain the same.
Explanation:
IMPORTANT: The solution to the question (a) and (b) that is (a) Draw a neat snapshot mode labeled vector picture of the wave. (b) Draw a neat movie mode labeled vector picture of the wave is there in the ATTACHED FILE/PICTURE.
It is also worthy of note to know that in anything Electromagnetic wave, the magnetic field, the Electric Field and their direction of propagation are perpendicular to each other.
Therefore, knowing the fact above we can say that in yellow light, the magnetic field is in the y-direction and the Electric Field is in the z-direction.
Hence, the solution to option C is given below;
(C).If the wave were to represent blue light instead of yellow light, picture (a) will remain the same because both light are Electromagnetic wave, although the wavelength will have to change. Picture (b) will also remain the same because they are both Electromagnetic waves and possess similar properties.
