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].(adsbygoogle = window.adsbygoogle || []).push({});

(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

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Gauss' Law and magnitude of the electric field

**Physics Forums | Science Articles, Homework Help, Discussion**