Resistance: Non-constant Surface Area

In summary, the problem reads:A circular disk of radius r and thickness d is made of material with resistivity p.The resistance between the points a and b (a is the front of the curved side of the cylinder and b is the backside) is independent of the radius and is given by R = [pi]p/2d.
  • #1
discoverer02
138
1
I think I'm on the verge of a breakthrough on this problem, but it's just not coming.

Please tell me where my approach goes wrong or whether I'm correct but don't know how to integrate the resulting equation properly.

The problem reads: A circular disk of radius r and thickness d is made of material with resistivity p. Show that the resistance between the points a and b (a is the front of the curved side of the cylinder and b is the backside) is independent of the radius and is given by R = [pi]p/2d.

I start out with R = pL/A where L is length and A is the surface area perpendicular to the direction of the flow of the charges.

dR = pdx/A

A = d * the width of the rectangular surface area for the little sliver of the cylinder, which is given by the equation of the circle y = 2[squ] (r^2-(x-r)^2).

Is this correct?

I then integrate R = p/2d[inte] dx/[squ](r^2-(x-r)^2) from 0 to 2r?

This gives me R= [pi]pr^2/8d.

Obviously wrong!

Any suggestions would be greatly appreciated.

Thanks.
 
Physics news on Phys.org
  • #2
Originally posted by discoverer02
Is this correct?

I am a little fuzzy on where a and b are, so I don't know if your setup is correct, but...

I then integrate R = p/2d[inte] dx/[squ](r^2-(x-r)^2) from 0 to 2r?

This gives me R= [pi]pr^2/8d.

I don't know how you did that integral, but when I do it I get exactly what you say I am supposed to get.

R=(ρ/2d)∫dx(r2-(x-r)2)-1/2

from 0 to 2r.

Let u=x-r
so du=dx

then...

R=(ρ/2d)∫du/(r2-u2)-1/2

from -r to r.

I get:

R=(ρ/2d)arcsin(u/r)|-rr

When you evaluate that, you will get R=ρπ/2d, as advertised.
 
  • #3
It's been a year since I've taken my last calculus class so I need to brush up. I used a table of integrals and I might have misread something.

I'm sorry that I didn't include a diagram. Imagine the cylinder sitting in front of you on its flat side. a to b is the front side to the back side or left side to right side. In this case the width of the rectangular surface area increases as you move from a to b according to the equation of a circle of radius r centered at (r,0).

So it looks like I set the integral up correctly, but evaluated it incorrectly.

Thanks so much for your help. It came just in the nick of time.:smile:
 
  • #4
Originally posted by discoverer02
It's been a year since I've taken my last calculus class so I need to brush up. I used a table of integrals and I might have misread something.

That must be it. As for me, I just happen to be teaching Calculus II this semester, and I taught my class that exact integral last week, so it is still pretty fresh in my mind. If not for that, I probably would have gone to an integral table myself!
 
  • #5
I suspect that your problem is that you "lost track" of the fact that the square root is in the denominator. The way I would do this, by the way, is integrate from -r to r rather than 0 to 2r with the origin at the center rather than in at p. That precisely the same as making the substitution "u= x-r" in your integral and gives

int(u= -r to r) (r^2- u^2)^(-1/2) du

A standard way to integrate something like that is to make the substitution u= r sin[theta]. Then du= r cos[theta] d[theta] and
r^2- u^2= r^2(1- sin^2[theta])= r^2 cos^2[theta] so that the square root gives r cos[theta]. Because those are in the DENOMINATOR, they will precisely cancel the "r cos[theta]" in the numerator. Your integral becomes int([theta]= -[pi/2] to [pi]/2) d[theta]= [pi].

That gives the result you want.
 
  • #6
Thanks for tip.

I'm sure I'll plenty of use for it during this semester's Electricity and Magnetism class.
 

1. What is resistance?

Resistance is the measure of an object's ability to resist the flow of electric current. It is represented by the symbol "R" and is measured in ohms (Ω).

2. How does resistance change with non-constant surface area?

The resistance of an object with non-constant surface area will also change. This is because as the surface area changes, the path for the electric current to flow also changes, resulting in a different resistance value.

3. What factors affect resistance in non-constant surface area?

The main factor that affects resistance in non-constant surface area is the shape and size of the object. Objects with larger surface areas will have lower resistance compared to objects with smaller surface areas.

4. How can resistance be calculated for non-constant surface area?

The formula for calculating resistance in non-constant surface area is R = ρ (L/A), where ρ is the resistivity of the material, L is the length of the object, and A is the cross-sectional area. This formula takes into account the changes in surface area and can be used to calculate resistance for different objects.

5. How does resistance impact the flow of electric current in non-constant surface area?

In non-constant surface area, resistance can impact the flow of electric current by either increasing or decreasing it. Objects with higher resistance will impede the flow of current, while objects with lower resistance will allow for a smoother flow of current.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
631
  • Introductory Physics Homework Help
2
Replies
63
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
779
Replies
13
Views
822
  • Introductory Physics Homework Help
Replies
4
Views
1K
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
139
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
Back
Top