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

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​

 

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]