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The example

In the textbook (Modern Physics for Scientists and Engineers by Douglas Giancoli) it asks for a capacitance calculation. However, I want just the explanation of the electric field calculation between two concentric cylinders. Why does he ignore the outer cylinder and write the electric field only due to the inner one?

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  • $\begingroup$ What is the electric field due to the outer cylinder at distance $<R_a$? (Hint: use Gauss' law) $\endgroup$
    – Sha Vuklia
    May 21, 2017 at 22:35
  • $\begingroup$ @ShaVuklia Oops, yes definitely it is 0 due to the oiter cylindre. Sorry for the stupidity:( $\endgroup$ May 21, 2017 at 22:40
  • $\begingroup$ No worries, it was a good question! It shows you were reading the text critically - and that's good! $\endgroup$
    – Sha Vuklia
    May 21, 2017 at 23:09
  • $\begingroup$ Please don't post questions that consist of cell phone pictures from a book: physics.meta.stackexchange.com/q/10563 . This severely limits the usefulness of the question, e.g., by making it not searchable in search engines and making it inaccessible to visually impaired people. Also, I'll add the homework-and-exercises tag, but in the future please use that tag on questions of this type. $\endgroup$
    – user4552
    May 27, 2018 at 18:39

1 Answer 1

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It is a property of the electrostatic force (and actually all inverse square force laws) that the amount of field passing through a Gaussian surface is dependent only on the amount of charge within the surface. This is evident in the expression for the flux through a closed Gaussian surface

$\Phi_{E} = \frac{q_{in}}{\epsilon_{0}} $

Thus if we were to calculate the flux through a cylinder whose radius falls between the two charged cylinders, the calculation would be the same, regardless of the existence of exterior charged cylinder.

Now while the flux through the Gaussian surface will remain unchanged regardless of any exterior charges, the exact expression for the field does not necessarily always remain the same. Thus we must still use the symmetries of the total charge in the particular problem to determine the particular geometry of the field.

Since the total charge distribution has a cylindrical symmetry, we can intuit that the field will also have cylindrical symmetry, and thus must only be dependent on the distance from the central axis.

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