Electric field of a non uniform charge of a cylinder

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  • #1
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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 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
 

Answers and Replies

  • #2
TSny
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Hello.

The formula you used in part (c) for E is not valid for this problem.

Try Gauss' law to derive E inside the cylinder.
 
  • #3
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like ∫(Ar^2)*r*dr = A∫r^3*dr
then plug everything back to the original equation?
 
  • #4
TSny
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like ∫(Ar^2)*r*dr = A∫r^3*dr
then plug everything back to the original equation?
Not sure I'm following you here. Does your integral represent charge in some region? If so, what region?

What equation is the "original equation"?

Since this looks like a problem for the application of Gauss' law, start with the statement of Gauss' law. Then think about how you would apply this law to the problem of finding E inside the cylinder.

The formula you used in part (c) for E is for the case where the charge density is uniform inside the cylinder. If you look back in your notes or textbook at the derivation of this formula using Gauss' law, it should provide a guide for setting up Gauss' law for your problem.
 
  • #5
BvU
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Almost: like ##\int_0^r \rho dV = 2\pi A\int_0^r r'^2 r'dr'##

[edit]drat, crossed TS reply again. Over to you, TS!

And: I take it the "equation, electric field inside the cylinder" is for a homogeneous charge distribution, which we don't have here. Right, Twin ?
 
  • #6
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i got it, thanks....so the density charge gets higher as the radius get bigger.
so i integrated the parts Qenc = ∫(A*r^2)2∏*r*H*dr r=Radius of gauss
put into the gauss law
EA=Qenclosed/εo

E(2∏*r*H)=(∫(A*r^2)2∏*r*H*dr)/εo

rearrange everything and cancel some of the variable (jump to final equation)

E=(A*r^3)/(4*εo)
 
  • #7
TSny
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Nice. Good work!
 

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