How Do You Calculate Average Charge Density in a Cylinder?

AI Thread Summary
To calculate the average charge density in a cylinder with a charge distribution of ρ = Cr², one must integrate the charge density over the volume of the cylinder. The total charge Q can be found using the volume integral in cylindrical coordinates, specifically integrating ρ(r) dV. The integration involves the limits for radius (0 to a), angle (0 to 2π), and length (0 to L), resulting in the expression Q = 4CπR⁵ / 5. The average charge density is then derived by dividing the total charge by the volume of the cylinder. This method ensures accurate calculation of the average charge density within the specified parameters.
Roodles01
Messages
125
Reaction score
0

Homework Statement


A cylinder of radius a and length l has charge distribution

ρ=Cr2
where C is a constant and r is radial distance in cylindrical coordinates.
Derive an expression for the average charge density within the cylinder.

Homework Equations


Well, charge density given is within the volume, I think.
So for a point on the axis of the cylinder should be ρ divided by the length shouldn't it? Or is that being too simple?

The Attempt at a Solution



ρ = Cr2 / l
[/B]
Can someone confirm this or point me in the right direction, please.
 
Physics news on Phys.org
First you need to integrate the distribution over the cylinder to find the total charge.
 
∫ Cr2 dr

C ∫ r2 dr (0 < l < L)

CL3/3
 
The radius 0 < r < a .
The length L.

You need to integrate over the VOLUME of the cylinder to find the total charge in the volume. I advise working in cylindrical co-ordinates. where dV = rdrd\theta dz
 
s ρ(r) dV = Q

∫ Cr2 * 4πr2 dr = 4Cπ ∫ r4 dr

so
Q = 4CπR5 / 5
 
You need to integrate over the cylinder.

\int dV \ = \iiint rdrd\theta dz\ = \int_{0}^{L} dz \ \int_{0}^{2\pi}d\theta \int_{0}^{a}rdr This is the volume integral for a cylinder and as you can see, doing the integral gives the volume of a cylinder of radius, a, and length, L. But since your integrating a function over this volume, you want
\int \rho(r) dV You can split the integral similarly to find the total charge.
 
Last edited:
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top