Griffith's book on electrodynamics says the electric field between 2 coaxial metal tubes of charge $+q$ and $-q$ is found by using Gauss's law, where the gaussian surface is a cylinder with radius between the outer and inner tube. But that doesn't take into account the electric field created by the outer tube, it will only find the elextric field created by the inner tube, Gauss's law only take into account the enclosed charge.

So my question is, shouldn't the electric field between the 2 tubes be a superposition of the electric field created by each tube?


Gauss's law always holds. Imagine for a moment that the inner tube is the only tube present. Then, using a symmetry argument, you can calculate the electric field using Gauss's law. Now imagine adding some kind of charge distribution outside of your Gaussian volume. While the total electric flux through the boundary of your volume remains unchanged, its distribution might change. In other words, it might go up in some places and go down in other places. So the symmetry argument no longer works, because the charge distribution outside the Gaussian volume is no longer symmetrical.

Fortunately, in your case the outer charge distribution is symmetrical in exactly the same way that the inner charge distribution is. So the symmetry argument still goes through. In fact, you can do the calculation with only the outer tube, in which case you find that the electric field inside the tube is 0. So the electric field for the space between the tubes is the sum of the electric field contributed by each tube, just as expected.


in gauss law when you write $\displaystyle\oint E.dS=\dfrac{Q_{tot.}}{\epsilon_{0}}$ ,

then $Q_{tot}$ is the total charge enclosed by gaussian surface

but that $\vec E$ i.e, electric field is not only due to charges present inside the gaussian surface

but is due to contribution of both the charges inside gaussian surface + charges outside gaussian surface as well

so even if you choose gaussian surface in between inner and outer radius of coaxial cable the charge contribution comes only through inner tube but electric field at the intermediary location between inner and outer radii is due to contribution of both inner tube +outer tube as well


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