Electric Field Inside Cylindrical Capacitor

[Idontknowhatimdoing]

Homework Statement

When we want to find the electric field inside of a cylindrical capacitor, we can use Gauss's law and the relation between flux and field to calculate what this field is.

Relevant Equations

Gauss's law

we know that flux is equal to the area integral of electric field dotted with dA and we can set this equal to charge enclosed divided by epsilon naught. Thus, in this case, the integral simplifies to E * A = (q_enclosed)/(ε_naught) when we choose a cylindrical gaussian surface with radius of r.

My question, then, is why are we allowed to use this relation to find electric field inside of the capacitor. I thought that the electric field that is calculated only describes the electric field that the charges enclosed produce. In this situation, wouldn't the negative charges on the outside cylinder affect the Electric Field too? If it does, the electric field that we find wouldn't be the right electric field between the cylinders. Is there a reason that we can ignore the negative charge on the outside cylinder in this capacitor? Or is my thought process incorrect?

 

[kuruman]

Look at the drawing on the right. Whatever field lines leave the inner surface where the positive charges are located terminate at the inner surface of the outer cylinder where the negative charges are. The electric flux through the dotted Gaussian surface essentially counts all the electric field lines. If you were to somehow account for the charges outside the Gaussian surface, you will be double-counting.

 

[nasu]

@OP. What do you think is the magnitude of the field produced by the outer charge inside the cylindrical shell?