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## Homework Statement

Consider a long, uniformly charged, cylindrical insulator of radius R and length l with charge density 1.8x10^-6 C/m^3. The value of the permittivity of space is 8.85419 × 10^−12.

What is the magnitude of the electric field inside the insulate at a distance 1.3cm from the axis (1.3cm < R)?

## Homework Equations

[tex]\phi = \int{EdA} = q_{inside}/\epsilon_{0}[/tex]

## The Attempt at a Solution

[tex]\int{EdA} = q_{inside} = [/tex] rho*V

[tex]\int{EdA} = q_{inside} = [/tex] rho*pi*r

^{2}*l

Due to symmetry, E must be constant and can be pulled outside the integral. Unfortunately, I get stuck at this step. I'm having trouble imagining the field and determining what the integral of dA actually is? I know it's not the complete surface area of the cylinder, but I'm unsure what it is.

Any help is greatly appreciated!

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