# Electric field due to infinite line of charge [closed]

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 gerbilJan 26 '18 at 0:35

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• The sentence beginning "By symmetry, the magnitude $E$..." is only true for an infinite cylinder, not for a finite one. – Chris Jan 23 '18 at 5:17
• Still like if we apply the same formula on a finite linear charge won't the result be same considering that length of the linear charge and height of the cylinder will cancel each other out? – user180358 Jan 23 '18 at 5:19
• – sammy gerbil Jan 26 '18 at 0:35
• I don't see why this was closed as a "homework-like question", though I agree that it's a duplicate of the one linked by @sammygerbil. – Michael Seifert Jan 26 '18 at 0:47
• @MichaelSeifert More than one reason was given by the 5 of us voted to close the question. Only the majority reason is given to explain why the question has been put on hold. In the event of a tie the last vote cast is the decider. See close vote reason logic when there isnt a majority – sammy gerbil Jan 26 '18 at 1:07

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 $\int\vec E\cdot d\vec A\ne EA$.