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: 4 times grease
Explanation: Force F= C · q1·q2/r². C = Coulomb's constant.
If charges double you have 2q1 and 2q2.
Answer:
Hi, Since energy levels in the nucleus are much higher than those in the gas, the nucleus will cool down by emitting a more energetic electromagnetic wave called a gamma ray . Gamma ray emission causes no change in the number of particles in the nucleus meaning both the atomic number and mass number remain the same.
Explanation :
It is given that,
Diameter of the coil, d = 20 cm = 0.2 m
Radius of the coil, r = 0.1 m
Number of turns, N = 3000
Induced EMF, 
Magnitude of Earth's field, 
We need to find the angular frequency with which it is rotated. The induced emf due to rotation is given by :
So, the angular frequency with which the loop is rotated is 159.15 rad/s. Hence, this is the required solution.
