- #1
nolachrymose
- 71
- 0
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*r2*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!
Last edited: