Electric Potential for Concentric Cylinders

[mrshappy0]

Homework Statement

What is V(P) – V(R), the potential difference between points P and R? Point P is located at (x,y) = (50.0 cm, 50.0 cm).

Homework Equations

V=-∫Edl
flux=∫EdA=Q/ε

The Attempt at a Solution

My attempt is in the attached photo to illustrate that I have worked the problem out and keep getting the wrong answer. I was able to obtain the electric field at point P but not V(P)-V(R). I assumed that you would allow rd to be x and dl to be dx. When it comes to plugging in the numbers something isn't lining up so I figured it is must be conceptual.

 

[mrshappy0]

excuse me that r10 in the image should be ra

 

[rude man]

You need to state the problem. Exactly as given to you.

 

[mrshappy0]

I did state the problem. It's right in the section that asked to state the problem. I am not looking for numbers. Just to be extra sure for you: An infinitely long solid insulating cylinder of radius a = 4.5 cm is positioned with its symmetry axis along the z-axis as shown. The cylinder is uniformly charged with a charge density ρ = 49.0 μC/m3. Concentric with the cylinder is a cylindrical conducting shell of inner radius b = 13.6 cm, and outer radius c = 17.6 cm. The conducting shell has a linear charge density λ = -0.53μC/m.

Actual question:What is V(P) – V(R), the potential difference between points P and R? Point P is located at (x,y) = (50.0 cm, 50.0 cm).


CORRECTION: image shows P and R labels mixed.

 

[mrshappy0]

I really am confident that I integrated correctly because the electric field expression is correct.

 

[haruspex]

When you integrate a field along a path, you have to be aware that the field and the distance element are both vectors. The integral is not ∫|E|.|dx| but ∫E.dx, i.e. the dot product. So you need to determine the component of the field in the direction of the distance element. Did you do that?

 

[rude man]

I did state the problem. It's right in the section that asked to state the problem. I am not looking for numbers. Just to be extra sure for you: An infinitely long solid insulating cylinder of radius a = 4.5 cm is positioned with its symmetry axis along the z-axis as shown. The cylinder is uniformly charged with a charge density ρ = 49.0 μC/m3. Concentric with the cylinder is a cylindrical conducting shell of inner radius b = 13.6 cm, and outer radius c = 17.6 cm. The conducting shell has a linear charge density λ = -0.53μC/m.

Actual question:What is V(P) – V(R), the potential difference between points P and R? Point P is located at (x,y) = (50.0 cm, 50.0 cm).


CORRECTION: image shows P and R labels mixed.

And where is R located? Is it (0, 50cm)?

 

[mrshappy0]

haruspex - Yes, I did that.

rude man - Yes, it is (0, 50cm). Sorry for the missed info but again when I posted the question I wasn't asking for the numbers I was just looking for some feedback on my integral that I set up and the use of the Electric field expression.

I have discovered that the expression is correct and so is the integral.

 

[haruspex]

The statement of the problem is not as clear as you seem to think. You show three circles. The outer two are the walls of an infinite cylinder, right? What about the one radius a?
Pls post your working for whichever part is still not working out correctly. (All your working would be better still.)

Just realized the title says concentric cylinders, but then I'm puzzled as to why the two contributions to the field look so different, with an r_a² term in one but no corresponding term in the other. Since it's constant, I guess it doesn't matter.

 

[rude man]

haruspex - Yes, I did that.

rude man - Yes, it is (0, 50cm). Sorry for the missed info but again when I posted the question I wasn't asking for the numbers I was just looking for some feedback on my integral that I set up and the use of the Electric field expression.

I have discovered that the expression is correct and so is the integral.

OK, I agree the way to do this problem is to integrate E*ds where ds is a vector element pointing from R to P.

I was just thinking - if you moved P behind R so that it is the same distance from the axis as before, would its potential not be unchanged? And would that not immensely simplify the integration of E*ds from R to P?

 

[haruspex]

if you moved P behind R so that it is the same distance from the axis as before, would its potential not be unchanged?

Or even, move it to (50/√2, 50/√2). Or is that what you meant?

 

[rude man]

Or even, move it to (50/√2, 50/√2). Or is that what you meant?

P is at (50,50) and so is 50√2 away from the axis (perpendicular distance). If you move P to (0,50√2) on the y-axis it would be behind R and in a straight line P to R to the axis. P would still be the same perpendicular distance from the axis as before, so its potential would not change.

Now you have an easy integration along the y-axis from R to P.

 

[haruspex]

P is at (50,50) and so is 50√2 away from the axis (perpendicular distance).

ok, I didn't see the statement that P and R were labelled wrongly.

 

[rude man]

P is at (50,50) and so is 50√2 away from the axis (perpendicular distance). If you move P to (0,50√2) on the y-axis it would be behind R and in a straight line P to R to the axis. P would still be the same perpendicular distance from the axis as before, so its potential would not change.

Now you have an easy integration along the y-axis from R to P.

EDIT: Well, time to correct myself again.

The potential of P if moved to a new position but at the same distance from its present position perpendicular to the axis is not the same. That's because infinite current lengths have infinite potential as measured the usual way (from infinity).

BUT - it's still the right idea:

Move P along a constant - radius (50√2) circle until it's behind R on a perpendicular to the axis as I said before. That's one integration.

Then integrate again from P to R along the radial distance between them. Then put a - sign on the sum of the two integrals since we want V(P) - V(R), not V(R) - V(P).

You then have two very easy integrations to perform (hint: neither requires integral calculus!).

 

[mrshappy0]

EDIT: Well, time to correct myself again.

The potential of P if moved to a new position but at the same distance from its present position perpendicular to the axis is not the same. That's because infinite current lengths have infinite potential as measured the usual way (from infinity).

BUT - it's still the right idea:

Move P along a constant - radius (50√2) circle until it's behind R on a perpendicular to the axis as I said before. That's one integration.

Then integrate again from P to R along the radial distance between them. Then put a - sign on the sum of the two integrals since we want V(P) - V(R), not V(R) - V(P).

You then have two very easy integrations to perform (hint: neither requires integral calculus!).

So with correct R,P labels - sorry, again for the mix up:

R being (0,50), I assumed anywhere at a fixed R=50 it will have the same potential around the cylinder. The same for P (50,50), knowing that it's R was 50√2.

So yes, I should have been a lot more clear. It's time consuming posting clear problems on here but I'll make more effort.

Anyways, I did the integral with boundaries from P and R, with everything constant coming out in front and 1/x dx remaining. Thus a simple integral. This ended up working out to the right answer, as I originally thought.

I made calculator error. Which is why I don't like automated HW problems- HUGE WASTE OF TIME!

Thanks for the help!

 

[rude man]

So with correct R,P labels - sorry, again for the mix up:

R being (0,50), I assumed anywhere at a fixed R=50 it will have the same potential around the cylinder. The same for P (50,50), knowing that it's R was 50√2.

I thought so too for a while but that is wrong. The problem as I said has to do with infinities. The potential of a point near the infinite wire is unfortunately infinity so you can't subtract one infinity from another .. which is why the right way was to integrate the electric field, which I see you've done correctly.

Cheers!