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:
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Answer:
Troposphere
High-pressure areas form due to downward motion through the troposphere, the atmospheric layer where weather occurs.
Answer:
Explanation:
The variables we know and are given are:
time, t = 20s
Charge, Q = 3x1-^-6 electrons, which is just 3x10^-6C (C stands for Coulombs, which is the unit for Charge)
We need to find the current, I, and since we know Q and t we can substitute these values into the given equation:
I=Q/t (which if you look at what the RHS is saying, its Charge over time, or more literally means the amount of charge passing a point over a period of time)
If we substitute these values, we will get I as:
I = Q / t
I = 3x10^-6 / 20
I = 1.5x10^-7 A
Hope this helps!
