# electric field of a non uniform charge of a cylinder

**1. The problem statement, all variables and given/known data**

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?

**2. Relevant 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)

**3. 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 in to 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, isnt it become voltage?

thank you for your time and reply

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