Why does the E-field inside a cylinder =0?

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SUMMARY

The electric field (E) inside a long, thin-walled metal tube carrying a surface charge density (λ) is zero for any point within the radius (R) of the tube. For points outside the tube (r > R), the electric field is given by the formula E = λ / (2πrε0). This conclusion is based on the principle that the electric field inside a closed conductor, devoid of internal charges, remains zero due to the inability of field lines to penetrate the conductor.

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Homework Statement



Figure 24-32 shows a section of a long, thin-walled metal tube of radius R, carrying a charge per unit length  on its surface. Derive expressions for E in terms of distance R from the tube axis, considering both (a) r > R and (b) r < R. Plot your results for the r = 0 to r = 5cm, assuming that  λ= 2*10^-8
C=m//and R = 3cm


[i can't put the figure since the question is in the book]

Homework Equations



q = ε0 ∫ E dA

The Attempt at a Solution



i got the answer to (a) E=λ/2∏rε0

i don't understand why (b) is 0
 
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You know that field inside a closed conductor, with no internal charges, is 0. You can think of this as being because field lines cannot pass through it. This tube is 'long'; as long as you are not near the ends, any field reaching in from outside will be small.
 

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