Gauss' Law and magnitude of the electric field

  • #1
DivGradCurl
372
0
A long, nonconducting, solid cylinder of radius 4.5 cm has a nonuniform volume charge density that is a function of the radial distance r from the axis of the cylinder, as given by [tex]\rho = A r^2[/tex], with [tex]A = 3.0 \mbox{ }\mu C/\mbox{m}^5[/tex].

(a) What is the magnitude of the electric field at a radial distance of 3.5 cm from the axis of the cylinder?

Gauss' Law:

[tex]\Phi _{\mbox{Net}} = \oint \vec{E} \cdot d\vec{S} = \int _{\mbox{Sides}} \vec{E} \cdot d\vec{S} = ES =\frac{Q_{\mbox{Enc}}}{\epsilon _0} [/tex]

Let a = 3.5 cm. Then:

[tex]E\left( 2\pi a L \right)=\frac{\rho \left( \pi a^ 2 L \right)}{\epsilon _0} \Longrightarrow E = \frac{\rho a}{2\epsilon _0} = \frac{Aa^ 3}{2\epsilon _0} \approx 7.3 \mbox{ } \frac{N}{C}[/tex]

This is wrong, but I don't know where.

(b) What is the magnitude of the electric field at a radial distance of 5.5 cm from the axis of the cylinder?

Gauss' Law:

Let b = 5.5 cm. Then:

[tex]\Phi _{\mbox{Net}} = \oint \vec{E} \cdot d\vec{S} = \int _{\mbox{Sides}} \vec{E} \cdot d\vec{S} = ES =\frac{Q_{\mbox{Enc}}}{\epsilon _0} [/tex]

[tex]E\left( 2\pi b L \right)=\frac{\rho \left( \pi R^ 2 L \right)}{\epsilon _0} \Longrightarrow E = \frac{\rho R^2}{2b\epsilon _0} = \frac{AR^ 4}{2b\epsilon _0} \approx 1.3 \times 10 \mbox{ } \frac{N}{C}[/tex]

This is wrong, but I don't know where.

Any help is highly appreciated
 

Answers and Replies

  • #2
cliowa
191
0
Dear thiago_j

I think the way you apply Gauss' Law in general is correct. The one thing that is wrong for sure is your calculation of the enclosed charge.
As you mentioned the charge density is a function of the radius. In your calculations you ignore this. You just take the volume (with radius a) and act as if the whole cylinder had the density it has at radius a, which is not the case (as you stated).
So, in order to calculate the charge enclosed you need to integrate!
Some hints on how to do this: In general, proceed as you always do with integration. Take a certain radius r, calculate the volume of the hollow cylinder with radii r and r+dr and calculate the charge on this infinitesimal hollow cylinder. Then do the integration over the whole volume needed.

I hope it was useful.
Best regards...Cliowa
 
  • #3
mukundpa
Homework Helper
524
3
thiago_j said:
[tex]E\left( 2\pi a L \right)=\frac{\rho \left( \pi a^ 2 L \right)}{\epsilon _0} \Longrightarrow E = \frac{\rho a}{2\epsilon _0} = \frac{Aa^ 3}{2\epsilon _0} \approx 7.3 \mbox{ } \frac{N}{C}[/tex]

[tex]Q_{\mbox{Enc}} [/tex] is not [tex]\rho \left( \pi a^ 2 L \right)[/tex]

it is

[tex] \int_0^r \ ( Ar^2 )2 \pi r dr L[/tex]
 
  • #4
DivGradCurl
372
0
How about now?

(a) What is the magnitude of the electric field at a radial distance of 3.5 cm from the axis of the cylinder?

Gauss' Law:

[tex]\Phi _{\mbox{Net}} = \oint \vec{E} \cdot d\vec{S} = \int _{\mbox{Sides}} \vec{E} \cdot d\vec{S} = ES =\frac{Q_{\mbox{Enc}}}{\epsilon _0} [/tex]

Let a = 3.5 cm. Then:

[tex]{Q_{\mbox{Enc}} = \int _0 ^a \left( Ar^2 \right) 2\pi r L \: dr = 2\pi LA \int _0 ^a r^3 \: dr = \frac{\pi LA a^ 4}{2}[/tex]

and so

[tex]E\left( 2\pi a L \right)= \frac{\pi LA a^ 4}{2\epsilon _0} \Longrightarrow E = \frac{A a^3}{4\epsilon _0} \approx 3.6 \mbox{ } \frac{N}{C}[/tex]

(b) What is the magnitude of the electric field at a radial distance of 5.5 cm from the axis of the cylinder?

Gauss' Law:

[tex]\Phi _{\mbox{Net}} = \oint \vec{E} \cdot d\vec{S} = \int _{\mbox{Sides}} \vec{E} \cdot d\vec{S} = ES =\frac{Q_{\mbox{Enc}}}{\epsilon _0} [/tex]

Let b = 5.5 cm and R = 4.5 cm. Then:

[tex]{Q_{\mbox{Enc}} = \int _0 ^R \left( Ar^2 \right) 2\pi r L \: dr = 2\pi LA \int _0 ^R r^3 \: dr = \frac{\pi LA R^ 4}{2}[/tex]

and so

[tex]E\left( 2\pi b L \right)= \frac{\pi LA R^ 4}{2\epsilon _0} \Longrightarrow E = \frac{A R^4}{4b\epsilon _0} \approx 6.3 \mbox{ } \frac{N}{C}[/tex]
 
  • #5
mukundpa
Homework Helper
524
3
Appears correct.
 

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