How Do Charge Distributions Affect Electric Potential in Physics Problems?

  • Thread starter Thread starter Bashkir
  • Start date Start date
  • Tags Tags
    Purcell
Click For Summary
SUMMARY

The discussion focuses on solving physics problems related to electric potential due to charge distributions, specifically problems 2.18 and 2.12 from a textbook. Problem 2.18 involves a hollow circular cylinder with a total charge Q uniformly distributed over its surface, requiring the calculation of potential differences along its axis. Problem 2.12 concerns a right triangle with a uniform surface charge, sigma, and the determination of potential at a vertex. Participants emphasized the importance of correctly defining variables, understanding limits of integration, and applying geometric principles to derive solutions accurately.

PREREQUISITES
  • Understanding of electric potential and charge distributions
  • Familiarity with calculus, particularly integration techniques
  • Knowledge of geometric relationships in triangles
  • Proficiency in LaTeX for mathematical expressions
NEXT STEPS
  • Study the concept of continuous charge distributions in electrostatics
  • Learn about the application of integration limits in physics problems
  • Explore geometric interpretations of electric fields and potentials
  • Practice LaTeX formatting for mathematical equations and expressions
USEFUL FOR

Students in physics courses, educators teaching electrostatics, and anyone interested in mastering the mathematical aspects of electric potential and charge distributions.

Bashkir
Messages
31
Reaction score
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,

dq=Q\frac{dA/A}=\frac{Q/l}

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

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

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 I've 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.

dq=\sigma da

for da I called it,

da=\frac{1/2}ydx

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

\tan\frac{y/x}=\tan\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?
 
Last edited:
Physics news on Phys.org
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.
 
Bashkir said:
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,
dq=Q\frac{dA}{A}=\frac{Q}{l}
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
\phi=\int\frac{dq}{r}=\frac{Q}{b}\int\frac{dx}{r} 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 I've 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.
dq=\sigma da for da I called it,
da=\frac{1}{2}y\,dx
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
\tan\frac{y}{x}=\tan\frac{a}{b}
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?
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
3K
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 13 ·
Replies
13
Views
6K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 17 ·
Replies
17
Views
3K
Replies
4
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
19
Views
3K