Welcome to our community

Be a part of something great, join today!

Physics Cylindrical Capacitor formulae- forming an intuition

Quintessential

New member
Feb 3, 2014
7
So far I have learned about Coulomb's law, the electric field, gauss's law, the electric potential and now capacitance.

I feel that although I "know of" these topics, I don't actually "flow with them".
Ignoring the math for a second; I want to form an understanding. And I think calculating the capacitance formula of a cylindrical capacitor may help.




* From that picture I understand that the center cylinder has a positive value, and the outer cylinder has an equal and opposite value.

* There must be an electric field in midst of the cylinders flowing from the positive to the negative.

* (I'm still confused with the electric potential portion, but here goes) There is also an electric potential tied together with this field. And to find this potential \(\displaystyle V\), I must find the Electric Field \(\displaystyle E\).

* I know: \(\displaystyle \oint_S {E_n dA = \frac{1}{{\varepsilon _0 }}} Q_{inside}\)

* Using the given gaussian surface on that picture, I should find \(\displaystyle E\).

* Once I find \(\displaystyle E\), I should plug it into the formula: \(\displaystyle \Delta V=-Ed\) and \(\displaystyle d\) being the distance \(\displaystyle (b-a)\). Because, (...not certain) the electric field multiplied by a given distance equals the electric potential for that given distance.

* Now that I have \(\displaystyle \Delta V\) I use the following: \(\displaystyle C=\frac{Q}{\Delta V}\). I know the electric potential and the charge is \(\displaystyle Q\). With that, I find the capacitance.

I haven't done any calculations to prove this. I only wanted to know if my intuition was legal.

Thank You.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,884
So far I have learned about Coulomb's law, the electric field, gauss's law, the electric potential and now capacitance.

I feel that although I "know of" these topics, I don't actually "flow with them".
Ignoring the math for a second; I want to form an understanding. And I think calculating the capacitance formula of a cylindrical capacitor may help.


Hey Quintessential! Nice picture! ;)
(I like pictures. :eek:)

This is all correct with 1 caution and 1 exception, which I'll comment on below.


* From that picture I understand that the center cylinder has a positive value, and the outer cylinder has an equal and opposite value.

* There must be an electric field in midst of the cylinders flowing from the positive to the negative.

* (I'm still confused with the electric potential portion, but here goes) There is also an electric potential tied together with this field. And to find this potential \(\displaystyle V\), I must find the Electric Field \(\displaystyle E\).

* I know: \(\displaystyle \oint_S {E_n dA = \frac{1}{{\varepsilon _0 }}} Q_{inside}\)
Caution: this only applies for a static electric field in vacuum.


* Using the given gaussian surface on that picture, I should find \(\displaystyle E\).

* Once I find \(\displaystyle E\), I should plug it into the formula: \(\displaystyle \Delta V=-Ed\) and \(\displaystyle d\) being the distance \(\displaystyle (b-a)\). Because, (...not certain) the electric field multiplied by a given distance equals the electric potential for that given distance.
This is only true if the electric field is constant, which it is not in this case.
The proper formulas are:
$$\mathbf E = -\nabla V$$
$$\Delta V = - \int_a^b \mathbf E \cdot \mathbf {dl}$$


* Now that I have \(\displaystyle \Delta V\) I use the following: \(\displaystyle C=\frac{Q}{\Delta V}\). I know the electric potential and the charge is \(\displaystyle Q\). With that, I find the capacitance.

I haven't done any calculations to prove this. I only wanted to know if my intuition was legal.

Thank You.
 

Quintessential

New member
Feb 3, 2014
7
Thanks a bunch for the helpful input!

Regarding the following:

This is only true if the electric field is constant, which it is not in this case.
The proper formulas are:
$$\mathbf E = -\nabla V$$
$$\Delta V = - \int_a^b \mathbf E \cdot \mathbf {dl}$$
Had the cylinder been infinitely long vertically, would the electric field have been constant?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,884
Had the cylinder been infinitely long vertically, would the electric field have been constant?
No.
The electric field is diverging from the inner cylinder toward the outer cylinder.
That means it becomes weaker.

Note that electric field strength is proportional to the density of the drawn electric field lines.
Close to the inner cylinder the density of lines is higher than it is at the outer cylinder.
 

Quintessential

New member
Feb 3, 2014
7
Perfect. Makes sense.

\(\displaystyle \Delta V = -E \int_0^{b-a} {dl}\)

I think I can take E out of the dot product seeing as how \(\displaystyle cos(\theta)=1\), rather the Electric field lines are parallel with the normal of the inner cylinder surface packets \(\displaystyle dl\)

And I'll have to integrate from 0 to the distance between the cylinders, so: \(\displaystyle b-a\)

As for \(\displaystyle dl\), now what would that be?

\(\displaystyle \Delta V = -E 2 \pi al(b-a)\)

I'm going for the volume between the cylinders. Still not 100% on this...
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,884
Perfect. Makes sense.

\(\displaystyle \Delta V = -E \int_0^{b-a} {dl}\)

I think I can take E out of the dot product seeing as how \(\displaystyle cos(\theta)=1\), rather the Electric field lines are parallel with the normal of the inner cylinder surface packets \(\displaystyle dl\)
Neh. That won't work.
You can only bring $E$ outside of the integral if it is constant, but it's not.
It changes with the radius, so we might write $E=E(r)$, meaning that $E$ is a function of $r$.

Then we get:
$$\Delta V = -\int_a^b \mathbf E \cdot \mathbf{dl} = -\int_a^b E(r) dr$$

To find $E(r)$ you need to use the other formula
$$\oint_{r\text{ constant}} E(r) dA = \frac Q {\varepsilon_0}$$

When you have $E(r)$ you can integrate it to find $\Delta V$.