# Potential difference between a tube and a coaxial bar

• Karol
The electric potential at a place r is the work done by the electric field while a unit positive charge moves from r to the place of zero potential.Or:the electric field is negative gradient of the potential. So ##\int _{V_0}^{V(r)} dV=-\int _{r_0 }^r Edr' \rightarrow V(r)-V(r_0)=\int_r^{ r_0} Edr'##If V(r0)=0, you get ##V(r)= \int_r^{ r_0} Edr'##You can not put the zero of the potential at infinity in case of an infinite wire or...?The electric potential at a place r is the work

## Homework Statement

A capacitor made of a round solid bar and a thin round tube are coaxial, they are very long. the radii of the bar is ra and of the tube rb. they are both charged with λ, the charge per unit length, with opposite charges. what is the potential difference.

## Homework Equations

The field round a long wire: ##E=\frac{\lambda}{2\pi\varepsilon_0 r}##
The work to bring a charge to a point in a field:
$$W=\int_{r_a}^{r^b} E dr$$

## The Attempt at a Solution

The field round a solid round bar and a tube is equal to the field generated by a thin wire.
The potential due to the inner bar:
$$V_{ab}=\int_{r_a}^{r^b} E dr=\frac{\lambda}{2\pi\varepsilon_0}\int_{r_a}^{r^b}\frac{1}{r}dr=\frac{\lambda}{2\pi\varepsilon_0}\ln\left(\frac{r_b}{r_a}\right)$$
This is the answer in the book, but the field due to the pipe isn't taken in consideration, why not?
The potential inside the pipe, due to the pipe, is:
$$V=\frac{\lambda}{2\pi\varepsilon_0}\frac{1}{r_b}$$
The potential difference should be:
$$\Delta V=\frac{\lambda}{2\pi\varepsilon_0}\left(\ln\left(\frac{r_b}{r_a}\right)-\frac{1}{r_b}\right)$$

Karol said:

## Homework Statement

A capacitor made of a round solid bar and a thin round tube are coaxial, they are very long. the radii of the bar is ra and of the tube rb. they are both charged with λ, the charge per unit length, with opposite charges. what is the potential difference.

## Homework Equations

The field round a long wire: ##E=\frac{\lambda}{2\pi\varepsilon_0 r}##
The work to bring a charge to a point in a field:
$$W=\int_{r_a}^{r^b} E dr$$

## The Attempt at a Solution

The field round a solid round bar and a tube is equal to the field generated by a thin wire.
The potential due to the inner bar:

$$V_{ab}=\int_{r_a}^{r^b} E dr=\frac{\lambda}{2\pi\varepsilon_0}\int_{r_a}^{r^b}\frac{1}{r}dr=\frac{\lambda}{2\pi\varepsilon_0}\ln\left(\frac{r_b}{r_a}\right)$$
This is the answer in the book, but the field due to the pipe isn't taken in consideration, why not?

The electric field is negative gradient of the potential, so you get the potential difference between r and ra by integrating E between these limits.

$$V(r)-V_a = -\int_{r_a}^{r} E dr'=-\frac{\lambda}{2\pi\varepsilon_0}\int_{r_a}^{r}\frac{1}{r}dr'=-\frac{\lambda}{2\pi\varepsilon_0}\ln\left(\frac{r}{r_a}\right)$$
That is the potential difference between the bar and any point "r" between the bar and pipe. It has no sense to speak about potential inside the pipe "due to the pipe".

Karol said:
The potential due to the inner bar:
$$V_{ab}=\int_{r_a}^{r^b} E dr=\frac{\lambda}{2\pi\varepsilon_0}\int_{r_a}^{r^b}\frac{1}{r}dr=\frac{\lambda}{2\pi\varepsilon_0}\ln\left(\frac{r_b}{r_a}\right)$$
Correction in terminology: this is the potential difference between ra and rb, due to the inner bar.
This is the answer in the book, but the field due to the pipe isn't taken in consideration, why not?
The potential inside the pipe, due to the pipe, is:
$$V=\frac{\lambda}{2\pi\varepsilon_0}\frac{1}{r_b}$$
This is the potential at any location inside the pipe, due to the pipe. What is the potential difference between any two locations inside the pipe (e.g. at ra and rb), due to the pipe?

ehild said:
It has no sense to speak about potential inside the pipe "due to the pipe".
Why not? the pipe is charged.
If there was no bar inside, still you wouldn't speak of potential inside? of course you wood, so, why not now? why does the oppositely charged bar change anything?
Oh, now i saw jtbel's reply, and i understand that since the potential due to the pipe is constant inside it makes no difference

Karol said:
The potential inside the pipe, due to the pipe, is:
$$V=\frac{\lambda}{2\pi\varepsilon_0}\frac{1}{r_b}$$
Can be ##\frac{\lambda}{2\pi\varepsilon_0}\frac{1}{r_b}## potential? Watch out the dimension. It is the electric field due to the tube outside. What is the electric field inside?
What is the potential of a hollow tube inside and outside? Where do you put the zero of the potential?

The electric field inside is 0 and the potential of a hollow tube is ##V=\frac{Q}{4\pi\varepsilon_0}\frac{1}{r}##
The zero of the potential is in infinity

Karol said:
The electric field inside is 0 and the potential of a hollow tube is ##V=\frac{Q}{4\pi\varepsilon_0}\frac{1}{r}##
The zero of the potential is in infinity
No, it is not the potential of a hollow tube. It is the potential of a point charge or a sphere, at distance r from the centre.

How do you get the potential of a hollow tube at distance r from its axis?

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The potential at point a is the work done to bring a positive charge from infinity to a:
$$V(a)-V_\infty =V(a)= -\int_{\infty}^{r_a} E dr'=-\frac{\lambda}{2\pi\varepsilon_0}\int_{\infty}^{r_a}\frac{1}{r}dr'=-\frac{\lambda}{2\pi\varepsilon_0}\ln\left(\frac{r_a}{\infty}\right)=-\frac{\lambda}{2\pi\varepsilon_0}$$
It's not logical since it's in minus sign and to bring a positive test charge to a positive charge you need to invest work and the sign should be +, and secondly it doesn't depend on r.

Karol said:
The potential at point a is the work done to bring a positive charge from infinity to a:
$$V(a)-V_\infty =V(a)= -\int_{\infty}^{r_a} E dr'=-\frac{\lambda}{2\pi\varepsilon_0}\int_{\infty}^{r_a}\frac{1}{r}dr'=-\frac{\lambda}{2\pi\varepsilon_0}\ln\left(\frac{r_a}{\infty}\right)=-\frac{\lambda}{2\pi\varepsilon_0}$$
It's not logical since it's in minus sign and to bring a positive test charge to a positive charge you need to invest work and the sign should be +, and secondly it doesn't depend on r.
First, you can not write that ln(ra/∞)=1.
Second:a definite integral with respect to r does not depend on r. (both r-s in the integrand should be the same either with prime, or without it.)
Third: how is the electric potential defined?
You are right that it was not logical what you wrote :)

The electric potential at a place r is the work done by the electric field while a unit positive charge moves from r to the place of zero potential.
Or:
the electric field is negative gradient of the potential. So ##\int _{V_0}^{V(r)} dV=-\int _{r_0 }^r Edr' \rightarrow V(r)-V(r_0)=\int_r^{ r_0} Edr'##
If V(r0)=0, you get ##V(r)= \int_r^{ r_0} Edr'##

You can not put the zero of the potential at infinity in case of an infinite wire or cylinder.

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Is ##\int _{V_0}^{V(r)}## the work done by the field to bring a positive charge from r to the place of zero potential? if so, then it should be:
$$W_{r\rightarrow r_0}^{field}=\int_{V(r)}^{V_0}$$
no? it changes the direction and sign and becomes ##V(r_0)-V(r)##

Karol said:
Is ##\int _{V_0}^{V(r)}## the work done by the field to bring a positive charge from r to the place of zero potential? if so, then it should be:
$$W_{r\rightarrow r_0}^{field}=\int_{V(r)}^{V_0}$$
no? it changes the direction and sign and becomes ##V(r_0)-V(r)##
No. Read my post carefully. The work is negative potential difference.

So it's ##W_{r\rightarrow r_0}^{field}=-\int_{V(r)}^{V_0}dV=\int_{V_0}^{V(r)}dV=V(r)-V_0## ?

Karol said:
So it's ##W_{r\rightarrow r_0}^{field}=-\int_{V(r)}^{V_0}dV=\int_{V_0}^{V(r)}dV=V(r)-V_0## ?
Yes.

ehild said:
If V(r0)=0, you get ##V(r)= \int_r^{ r_0} Edr'##
You can not put the zero of the potential at infinity in case of an infinite wire or cylinder.

$$V(r)-V_a =-\frac{\lambda}{2\pi\varepsilon_0}\ln\left(\frac{r}{r_a}\right)$$
How do i find r0 where V(r0)=0?
I have to make the function $$\ln x=0\rightarrow x=1$$, and it happens when $$\frac{r_0}{r}=\frac{r_0}{r_0}$$ but it doesn't lead me anywhere
It's a round way. what i did is calculate V at r0 and it's 0, of course, but i can't find r0 since it's not isolated, like in the potential for a point charge:
##V(r)-V_a =\frac{Q}{4\pi\varepsilon_0}\left(\frac{1}{r}-\frac{1}{r_a}\right)##

You are free to choose the place where V=0. It can be ra, for example, that is, Va=0.

ehild said:
You are free to choose the place where V=0. It can be ra, for example, that is, Va=0.
$$V=-\int dV=-\int Edr'=-\frac{\lambda}{2\pi\varepsilon_0}\ln r+C$$
I want to find the constant C in order that the 0 potential energy will be in infinity:
$$V=0=-\frac{\lambda}{2\pi\varepsilon_0}\ln \infty+C$$
And it's no good

Karol said:
I want to find the constant C in order that the 0 potential energy will be in infinity:
It is not possible for an infinitely long cylinder.
The absolute potential has no Physical meaning. It is only the potential difference that counts. You were asked about the potential difference.

It's not possible because in ##V=-\frac{\lambda}{2\pi\varepsilon_0}\ln r## V gets infinitely higher and doesn't reach an asymptote like in gravitation: ##U=-\frac{GMm}{r}##? so i will have to choose a radius r where V=0?
So i choose ##r=2\cdot r_b## for V=0 then:
$$-\frac{\lambda}{2\pi\varepsilon_0}\ln (2r_b)+C=0\rightarrow C=\frac{\lambda}{2\pi\varepsilon_0}\ln (2r_b)$$
And the potential at rb, the face of the tube, is:
$$V(r=r_b)-V(r=2r_b)=V(r=r_b)=-\frac{\lambda}{2\pi\varepsilon_0}\ln (r_b)+\frac{\lambda}{2\pi\varepsilon_0}\ln (2r_b)=\frac{\lambda}{2\pi\varepsilon_0}\left(\ln (2r_b)-\ln r_b\right)=\frac{\lambda}{2\pi\varepsilon_0}\ln \frac{2r_b}{r_b}=\frac{\lambda}{2\pi\varepsilon_0}\ln 2$$
And this potential is in the pipe and doesn't make a difference to the potential due to the inner bar since it's constant.

You can choose r=rb for V=0, as well. And it does not matter as you need potential difference. To get the potential inside the pipe, you have to fit the potential "due to the bar" to the outside potential, so as the potential is continuous at r=rb.
There is a single potential U(r), which is a continuous function.

It has to be continuous in order to be unambiguous? then at r=r2b V due to the bar must also be 0, so in this case where the functions are the same the potential inside is the potential difference due to the pipe between rb and r2b, which inside is constant and equals ##=\frac{\lambda}{2\pi\varepsilon_0}\ln 2##
plus the potential difference between r and r2b due to the bar which is $$V=\frac{\lambda}{2\pi\varepsilon_0}\ln \frac{2r_b}{r}$$
And together:
##V_r=\frac{\lambda}{2\pi\varepsilon_0}\ln 2+\frac{\lambda}{2\pi\varepsilon_0}\ln \frac{2r_b}{r}##
##V_r=\frac{\lambda}{2\pi\varepsilon_0}\ln \frac{4r_b}{r}##

Be careful, the bar and the tube are oppositely charged.
You have different form of functions for inside the tube and outside it. The potential is continuous at r=rb (the outside surface of the cylinder), but the derivative dV/dr is not.)

I do not understand why do you stick to potential due to tube and due to to bar. There is only one potential function, given piecewise for ra<r<rb and r>rb. It is continuous everywhere, and its negative gradient is equal to the electric field.

I made a mistake, the continuity problem might be at rb, but since these 2 functions are the same, only with opposite signs, for the bar and the tube, then there is no continuity problem and inside the tube:
$$V_r=-\frac{\lambda}{2\pi\varepsilon_0}\ln 2+\frac{\lambda}{2\pi\varepsilon_0}\ln \frac{2r_b}{r}=\frac{\lambda}{2\pi\varepsilon_0}\ln \frac{r_b}{r}$$
For the combined system of the bar and the tube it doesn't make a difference where the 0 potential is since it will always come out ##V_r=\frac{\lambda}{2\pi\varepsilon_0}\ln \frac{r_b}{r}##, but for only the bar or the pipe it makes a difference, so how does it settle with the saying that the potential differences are the same no matter where the zero is?

The potential function contains an additive constant, according to the zero of the potential. The additive constant cancels if you take the potential difference between two points.

now i see i made a mistake. it has to be continuous at rb. since both functions are the same here, only with opposite signs it is obtained and the potential in the gap between the pipe and the bar is the sum of the potentials:
$$V_r=-\frac{\lambda}{2\pi\varepsilon_0}\ln 2+\frac{\lambda}{2\pi\varepsilon_0}\ln \frac{2r_b}{r}=\frac{\lambda}{2\pi\varepsilon_0}\ln \frac{r_b}{r}$$
If i change the zero point this result will still be true, but for each of them alone, the pipe or the bar, it will change, so how does it settle with the fact that only differences in potential have meaning and are the same?

Oh, i had, probably, a problem with the browser, now i see that i this question is registered, now i will read ehild's answer

When you have point charges, it has sense to calculate the potential as the sum of potential due to the single charges.
If you have problems with charged metal objects, like in this problem, the distribution of charges on the surfaces are influenced by the geometry of the set-up. In symmetric cases, it is easier if you determine the electric field , using Gauss' Law and integrate it to obtain the potential.

Look at the picture. In case of a single tube, the surface charge is on the outer surface. If the bar is placed in the middle, the charge of the tube is on the inner surface. Draw a Gaussian cylinder around the tube. The charge inside is zero, as the tube and the bar have equal and opposite charges. The electric field outside the cylinder is zero.
So
##V(R) - V(Ro) = -\int_{Ro}^R Edr = 0 ##.
The potential is constant. You can choose that potential equal to zero. (Ro can be at infinity, where it is naturally zero.

V(R)=V(Ro)=0) .That is true at Rb, too, so V(Rb) = 0.

Between the inner bar and the inner surface of tube the Gaussian cylinder encloses q charge. The electric field is

##E(R)=\frac{q}{\varepsilon 2\pi R L }##.

If the wall of the tube is very thin the radius of the inner surface is also Rb. As we know the potential there, we integrate from Rb to R to get the potential between Ra and Rb.

##V(R) - V(Rb) = -\int_{Rb}^R Edr = - \frac{q}{\varepsilon 2\pi L }\int_{Rb}^R (\frac{1}{r}dr)= \frac{q}{\varepsilon 2\pi L } \log( \frac {Rb}{R})##

The potential inside the metal bar is also constant and is equal to the potential at Ra.

As V(Rb)=0, the potential function is

V(R)=0 if R≥Rb;
##V(R)=\frac{q}{\varepsilon 2\pi L } \log( \frac {Rb}{R}) ## if Ra ≤ R≤ Rb;
##V(R)=\frac{q}{\varepsilon 2\pi L } \log( \frac {Rb}{Ra}) ## if R≤Ra.

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Thank you Ehild but i ask: if i only had the pipe, and decide that V=0 at ##r=2r_b##, for example, then the integration constant C is:
$$-\frac{\lambda}{2\pi\varepsilon_0}\ln (2r_b)+C=0\rightarrow C=\frac{\lambda}{2\pi\varepsilon_0}\ln (2r_b)$$
And the potential at rb:
$$V(r_b)=-\frac{\lambda}{2\pi\varepsilon_0}\ln (r_b)+\frac{\lambda}{2\pi\varepsilon_0}\ln (2r_b)=\frac{\lambda}{2\pi\varepsilon_0}\ln \frac{2r_b}{r_b}$$
$$V(r_b)=\frac{\lambda}{2\pi\varepsilon_0}\ln 2$$
And if i choose V=0 at ##r=3r_b## then ##V(r_b)=\frac{\lambda}{2\pi\varepsilon_0}\ln 3## and so on
So the choice of V=0 makes a difference to V, it shouldn't be

Karol said:
So the choice of V=0 makes a difference to V, it shouldn't be
The V(r) functions are different by an additive constant.
If V(2rb)=0
## V(r)=\frac{\lambda}{2 \pi \varepsilon_0} \ln (\frac{2r_b}{r})##
If 3rb=0

##
V(r)=\frac{\lambda}{2\pi\varepsilon_0}\ln( \frac{3r_b}{r})
##

But you can not measure V. You can only measure potential difference between two points. And then the additive constant cancels.
Determine the potential difference between d1 and d2:

With the first potential,
##
V(d1)=\frac{\lambda}{2\pi\varepsilon_0}\ln (\frac{2r_b}{d1})
##

and ##
V(d2)=\frac{\lambda}{2\pi\varepsilon_0}\ln (\frac{2r_b}{d2})
##
##
V(d2)-V(d1)=\frac{\lambda}{2\pi\varepsilon_0}\ln (\frac{d1}{d2}) ##, and it is the same with the second potential.

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Thank you very much, Ehild