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TwinCamGTS
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Homework Statement
A very long solid cylinder of radius R = 4.2 cm has a non-uniform volume charge density along its radial dimension, given by the function ρ = Ar2, where A = +2.2 µC/m5.
a)How much total charge is contained on a 1 m length of this cylinder?
b)Outside: What is the electric field at a radial distance of 5.2 cm from the axis of the cylinder?
c)Inside: What is the electric field at a radial distance of 3.2 cm from the axis of the cylinder?
Homework Equations
use gauss law to find the electric field
total charge = ρ * area of the cylinder = A*(r^2)*2∏*r*L = A*2∏*L*r^3
electric field inside the cylinder E = (ρ * r)/(2 * εo)
electric field outside the cylinder E = ( λ )/(2∏ * r * εo)
The Attempt at a Solution
a)How much total charge is contained on a 1 m length of this cylinder?
first of all we need to find the total charge of contained on a 1 m length of the cylinder
the volume charge density is changing by the radial dimension ( the radius).
so, to find the total charge for the cylinder with radius 4.2cm,
we need to do the integral for the charge equation
so total charge is
∫A*2∏*L*r^3 dr , the result is (.5∏*A*L*r^4)
plugin the value of A, L , and r we get (1.07532E-11 C)
Outside: What is the electric field at a radial distance of 5.2 cm from the axis of the cylinder?
electric field outside the cylinder E = ( λ )/(2∏ * r * εo) i got (3.71887 N/C)
c)Inside: What is the electric field at a radial distance of 3.2 cm from the axis of the cylinder?
I have problem with this one, based on the equation, electric field inside the cylinder
E = (ρ * r)/(2 * εo) , i just plug in everything into the formula
E = ((A*r^2)*r)/(2*εo)
= ((2.2E-6)*(.032^3))/(2*(8.85E-12))
= 4.07285 N/C
but the answe is wrong.
I just wonder, do i have to redo the integration for this part again? E = ((A*r^2)*r)/(2*εo)
I thought if i do the integration, isn't it become voltage?
thank you for your time and reply