• Support PF! Buy your school textbooks, materials and every day products Here!

Purcell, Chapter 2 Problems

  • Thread starter Bashkir
  • Start date
  • #1
31
1
These are more mathematics than they are problems with understanding of physics.
1. Homework Statement

The problems are 2.18 and 2.12 which are as follows,

2.18A hollow circular cylinder, of radius a, and length b, with open ends, has a
total charge Q uniformly distributed over its surface. What is the dierence in
potential between a point on the axis at one end and the midpoint of the eaxis?
Show by sketching some eld lines how you think the eld of this thing ought to
look.

2.12 The right triangle with vertex P at the origin, base b, and altitude a has a uniform surface charge, sigma.
Determine the potential at the vertex. First consider the contribution from a strip of width dx.

Homework Equations





The Attempt at a Solution


For 2.18, I arrived at the correct solution, but I am having trouble grasping a few concepts.

I started by calling the cylinder a set of segmented rings of thickness dx. The charge on each of these rings should be proportion to the total Q times the ratio of dA/A. or,

[tex]dq=Q\frac{dA/A}=\frac{Q/l}[/tex]

I reasoned that the radius of the cylinder was constant, so we were differentiating with respect to l. So far so good I think. Since it was a continuous charge distribution I wrote

[tex]\phi=\int\frac{dq/r}=\frac{Q/b}\int\frac{dx/r}[/tex]

I was stuck here and after looking for help online I was able to come to the limits of integration and then the rest of the problem became trivial.

I don't know how to Tex the limits of integration, but the upper bound and lower bounds were respectively b/2+x and -b/2+x

The limits of integration confuse me. The x is just some arbitrary distance on the x axis. What Ive gathered is, the upper bound basically gives you any point on the positive x axis, and then the lower bound gives you any point on the negative x-axis, even past the cylinder if you allow x to be a negative constant. Is this correct thinking? That was the only way I could rationalize it to myself, and if it is correct, how do I implore this type of analytical thinking? Is it really just practice and patter recognition? I don't think I would have ever come to those bounds. I would have attmped b/2 and -b/2 forever.

For 2.12 I am also having a little bit of mathematical trouble.

[tex]dq=\sigma da[/tex]

for da I called it,

[tex]da=\frac{1/2}ydx[/tex]

Y I solved for by using the fact that theta stays constant so

[tex]\tan\frac{y/x}=\tan\frac{a/b}[/tex]

Then I plugged that into the integral given for a continuous distribution of charge, calling r^2=x^2+y^2 and integrating dx from 0 to x. The answer I get is not what the book gives from the contribution from dx. Am I wrong in assuming a constant y?
 
Last edited:

Answers and Replies

  • #2
31
1
Ignoring the 2.12! After taking some time back to think, my change in area per strip is dxdy, and I hold DX constant while considering the contribution from it, and integrate dy/r from 0 to xa/b. That gets me pretty close, and then just integrating dx from 0 to x gives me the correct answer.
 
  • #3
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,580
1,196
These are more mathematics than they are problems with understanding of physics.
1. Homework Statement

The problems are 2.18 and 2.12 which are as follows,

2.18A hollow circular cylinder, of radius a, and length b, with open ends, has a
total charge Q uniformly distributed over its surface. What is the dierence in
potential between a point on the axis at one end and the midpoint of the eaxis?
Show by sketching some eld lines how you think the eld of this thing ought to
look.

2.12 The right triangle with vertex P at the origin, base b, and altitude a has a uniform surface charge, sigma.
Determine the potential at the vertex. First consider the contribution from a strip of width dx.

Homework Equations





The Attempt at a Solution


For 2.18, I arrived at the correct solution, but I am having trouble grasping a few concepts.

I started by calling the cylinder a set of segmented rings of thickness dx. The charge on each of these rings should be proportion to the total Q times the ratio of dA/A. or,
[tex]dq=Q\frac{dA}{A}=\frac{Q}{l}[/tex]
You need to define your variables. We can't read your mind or see what's on your paper. In any case, that last expression can't be right. It doesn't contain a differential. Also, the units don't work out.

I reasoned that the radius of the cylinder was constant, so we were differentiating with respect to l. So far so good I think. Since it was a continuous charge distribution I wrote
[tex]\phi=\int\frac{dq}{r}=\frac{Q}{b}\int\frac{dx}{r}[/tex] I was stuck here and after looking for help online I was able to come to the limits of integration and then the rest of the problem became trivial.
You're trying to calculate the potential at a point, right? What variable represents where the point is?

I don't know how to Tex the limits of integration, but the upper bound and lower bounds were respectively b/2+x and -b/2+x

The limits of integration confuse me. The x is just some arbitrary distance on the x axis. What Ive gathered is, the upper bound basically gives you any point on the positive x axis, and then the lower bound gives you any point on the negative x-axis, even past the cylinder if you allow x to be a negative constant. Is this correct thinking? That was the only way I could rationalize it to myself, and if it is correct, how do I implore this type of analytical thinking? Is it really just practice and patter recognition? I don't think I would have ever come to those bounds. I would have attmped b/2 and -b/2 forever.

For 2.12 I am also having a little bit of mathematical trouble.
[tex]dq=\sigma da[/tex] for da I called it,
[tex]da=\frac{1}{2}y\,dx[/tex]
I fixed your LaTeX for you. The way it rendered before, it was very misleading. You really need to proofread your post. Your expression for ##da## is wrong but at least now the units make sense.

Y I solved for by using the fact that theta stays constant so
[tex]\tan\frac{y}{x}=\tan\frac{a}{b}[/tex]
If you mean that ##y## is the height of the strip, then you don't want the ##\tan##s in there. It's just basic geometry of similar triangles: ##\frac{y}{x} = \frac{a}{b}##.

Then I plugged that into the integral given for a continuous distribution of charge, calling r^2=x^2+y^2 and integrating dx from 0 to x. The answer I get is not what the book gives from the contribution from dx. Am I wrong in assuming a constant y?
Since you figured the problem out, can you identify why your initial approach wouldn't work?
 

Related Threads on Purcell, Chapter 2 Problems

Replies
1
Views
4K
Replies
1
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
10
Views
5K
Replies
0
Views
4K
Replies
4
Views
1K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
3
Views
3K
Top