In my textbook it derives the electric field due to an infinitely long linear charge using a cylindrical Gaussian surface. However, if I take a finite-length linear charge, and draw the same cylinder, I get the same field. Why does the derivation only work for an infinitely long linear charge?
closed as off-topic by Mitchell, Chris♦, John Rennie, glS, sammy gerbil Jan 26 '18 at 0:35
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The sentence in your textbook beginning "By symmetry, the magnitude E..." is only true for an infinite cylinder, not for a finite one. Generally speaking, it is impossible to get the electric field using only Gauss' law without some symmetry to simplify the final expression.
You can't apply Gauss' law in any useful way for a finite line charge, because the electric field isn't normal to the surface of the cylinder, and so $\int\vec E\cdot d\vec A\ne EA$.