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

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The electric field (E) outside a long, thin-walled metal tube with surface charge density λ is given by E = λ/(2πrε0) for distances r greater than the tube's radius R. Inside the tube (for r < R), the electric field is zero because the field lines cannot penetrate the closed conductor, and there are no internal charges to create an electric field. This phenomenon is consistent with electrostatic principles, where the electric field inside a conductor in electrostatic equilibrium is always zero. The discussion emphasizes the importance of understanding how electric fields behave in relation to conductive materials. The conclusion reinforces that the electric field inside the cylinder is zero due to the nature of electric fields in conductors.
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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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