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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 $$V$$, I must find the Electric Field $$E$$.
* I know: $$\oint_S {E_n dA = \frac{1}{{\varepsilon _0 }}} Q_{inside}$$
* Using the given gaussian surface on that picture, I should find $$E$$.
* Once I find $$E$$, I should plug it into the formula: $$\Delta V=-Ed$$ and $$d$$ being the distance $$(b-a)$$. Because, (...not certain) the electric field multiplied by a given distance equals the electric potential for that given distance.
* Now that I have $$\Delta V$$ I use the following: $$C=\frac{Q}{\Delta V}$$. I know the electric potential and the charge is $$Q$$. 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.
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.

* 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 $$V$$, I must find the Electric Field $$E$$.
* I know: $$\oint_S {E_n dA = \frac{1}{{\varepsilon _0 }}} Q_{inside}$$
* Using the given gaussian surface on that picture, I should find $$E$$.
* Once I find $$E$$, I should plug it into the formula: $$\Delta V=-Ed$$ and $$d$$ being the distance $$(b-a)$$. Because, (...not certain) the electric field multiplied by a given distance equals the electric potential for that given distance.
* Now that I have $$\Delta V$$ I use the following: $$C=\frac{Q}{\Delta V}$$. I know the electric potential and the charge is $$Q$$. 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.