Answer: Ok, first lest see out problem.
It says it's a Long cylindrical charge distribution, So you can ignore the border effects on the ends of the cylinder.
Also by the gauss law we know that E¨*2*pi*r*L = Q/ε0
where Q is the total charge inside our gaussian surface, that will be a cylinder of radius r and heaight L.
So Q= rho*volume= pi*r*r*L*rho
so replacing : E = (1/2)*r*rho/ε0
you may ask, ¿why dont use R on the solution?
since you are calculating the field inside the cylinder, and the charge density is uniform inside of it, you don't see the charge that is outside, and in your calculation actuali doesn't matter how much charge is outside your gaussian surface, so R does not have an effect on the calculation.
R would matter if in the problem they give you the total charge of the cylinder, so when you only have the charge of a smaller r radius cylinder, you will have a relation between r and R that describes how much charge density you are enclosing.
Answer:
The magnetic flux through surface is 
Wb
Explanation:
Given :
Magnitude of magnetic field 
T
Radius of circle 
m
Angle between field and surface normal 
25°
From the formula of flux,
Where angle between magnetic field line and surface normal, 
area of circular surface.
Magnetic flux is given by,
Wb
Therefore, the magnetic flux through surface is Wb
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If the separation distance is doubled, then the electric field decreases by a factor of 4.
<h3>What is the electric field strength?</h3>
We know that the electric field strength is known to depend on the magnitude of the charge and the distance of separation. We know that the electric field refers to the region in which the influence of a charge is felt. Recall that a charge is a specie that is positively or negatively charged. The charge on a specie must always be shown by its sign.
We know that the electric field is the region in space where the influence of a charge can be felt. If a charge is placed in the vicinity of another charge, the second charge would experience a force due to the presence of the first charge. This is because the second charge was brought into the electric field of the first charge.
Thus we know that;
E = Kq/r^2
Where;
E = electric field strength
q = magnitude of charge
r = distance of separation
Now;
E = 9.0* 10^9 * 3.052 * 10^-6/(8.22 * 10^-2)^2
E = 4 N/C
Given that the electric filed strength is inversely proportional to the distance of separation, when the distance between the charges is doubled, the electric field decreases by a factor of 4.
Learn more about electric field strength:brainly.com/question/15170044?
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Answer:
E = h ν energy of electromagnetic particle
(b) has the greater frequency
