Chapter 3, Problem 3.15P

Volume charge density is located as follows; p_v=0 for p<1 mm and for p>2 mm, p_v=4p đ�œ‡C/m³ for 1 ) Calculate the total charge in the region 0 1,0 1<2mm, (b ) Use Gauss’s law to determine D_p at p= p_1. (c)Evaluate D _pat p=p_{1. (c) Evaluate} D _pa at p=0.8 mm,1.6.

To determine

(a)

Total charge in the given region.

Answer to Problem 3.15P

   $\frac{\text{8}\pi \text{L}}{\text{3}}\left[{\text{ρ}}_{\text{1}}^{\text{3}}{\text{-10}}^{\text{-9}}\right]\text{μC}$

Explanation of Solution

Given Information:

   $\begin{array}{c}{\rho }_v=0\text{ for }\rho <1\text{mm and }\rho >2\text{mm}\\ {\rho }_v=4\rho \mu {\text{ C/m}}^{\text{3}}\text{ for }1<\rho <2\text{mm}\end{array}$

Concept used:

   $\text{Q=}{\displaystyle \underset{\text{0}}{\overset{\text{L}}{\int }}{\displaystyle \underset{\text{0}}{\overset{\text{2π}}{\int }}{\displaystyle \underset{\text{0}\text{.001}}{\overset{{\text{ρ}}_{\text{1}}}{\int }}\text{4 ρ ρ dρ d}\phi {\text{ d}}_{\text{2}}}}}$

Calculation:

   $\begin{array}{c}\text{Q=}{{\displaystyle \underset{\text{0}}{\overset{\text{L}}{\int }}{\displaystyle \underset{\text{0}}{\overset{\text{2π}}{\int }}\left[\frac{{\text{4 ρ}}^{\text{3}}}{\text{3}}\right]}}}_{\text{0}\text{.001}}^{{\text{ρ}}_{\text{1}}}\text{d}\varphi \text{dz}\\ \text{=}{\displaystyle \underset{\text{0}}{\overset{\text{L}}{\int }}{\displaystyle \underset{\text{0}}{\overset{\text{2π}}{\int }}\left[\frac{{\text{4 ρ}}_{\text{1}}^{\text{3}}}{\text{3}}\text{-}\frac{{\text{4 × 10}}^{\text{-9}}}{\text{3}}\right]}}\text{ d}\varphi \text{dz}\\ \text{=}{\displaystyle \underset{\text{0}}{\overset{\text{L}}{\int }}\left[\frac{{\text{4 ρ}}_{\text{1}}^{\text{3}}\text{ × 2π}}{\text{3}}\text{-}\frac{{\text{4 × 2π × 10}}^{\text{-9}}}{\text{3}}\right]\text{ dz}}\\ \text{=}{\left[\left(\frac{{\text{4 ρ}}_{\text{1}}^{\text{3}}\text{× 2π}}{\text{3}}\text{-}\frac{{\text{4 × 2π × 10}}^{\text{-9}}}{\text{3}}\right)\text{Z}\right]}_{\text{0}}^{\text{L}}\\ =\frac{{\text{4 ρ}}_{\text{1}}^{\text{3}}\text{× 2π × }2}{\text{3}}\text{-}\frac{{\text{4 × 2π × 10}}^{\text{-9}}\text{ × L}}{\text{3}}\\ \text{=}\frac{\text{8πL}}{\text{3}}\left[{\text{ ρ}}_{\text{1}}^{\text{3}}{\text{ - 10}}^{\text{-9}}\right]\text{ μC}\end{array}$

Conclusion:

Total charge is $\frac{\text{8πL}}{\text{3}}\left[{\text{ ρ}}_{\text{1}}^{\text{3}}{\text{ - 10}}^{\text{-9}}\right]\text{ μC}$

To determine

(b)

The value of ${\text{D}}_{\text{ρ}}{\text{ at ρ = ρ}}_{\text{1}}$

Answer to Problem 3.15P

   ${\text{D}}_{\text{ρ}}{\text{(ρ}}_{\text{1}}\text{) = }\frac{{\text{4 (ρ}}_{\text{1}}^{\text{3}}{\text{ - 10}}^{\text{-9}}\text{)}}{{\text{3ρ}}_{\text{1}}}{\text{ μ C/m}}^{\text{2}}$

Explanation of Solution

Given Information:

   $\begin{array}{c}{\rho }_v=0\text{ for }\rho <1\text{mm and }\rho >2\text{mm}\\ {\rho }_v=4\rho \mu {\text{ C/m}}^{\text{3}}\text{ for }1<\rho <2\text{mm}\end{array}$

Concept used:

   $\text{Q =}\frac{\text{8πL}}{\text{3}}\left[{\text{ ρ}}_{\text{1}}^{\text{3}}{\text{ - 10}}^{\text{-9}}\right]\text{ μC}$ (Taken from part a)

Gauss's law, ${\text{Q = 2πρLD}}_{\text{ρ}}$

Calculation:

   $\begin{array}{c}{\text{D}}_{\text{Q}}\text{= }\frac{\text{Q}}{{\text{2πρ}}_{\text{1}}\text{L}}\\ \text{= }\frac{\text{8πL}\left({\text{ρ}}_{\text{1}}^{\text{3}}{\text{-10}}^{\text{-9}}\right)}{{\text{3 × 2πρ}}_{\text{1}}\text{L}}\\ \text{= }\frac{\text{4}}{\text{3}}\left({\text{ρ}}_{\text{1}}^{\text{3}}{\text{-10}}^{\text{-9}}\right){\text{ μC/m}}^{\text{2}}\end{array}$

Conclusion:

${\text{D}}_{\text{ρ}}{\text{ at ρ = ρ}}_{\text{1}}$ is $\frac{\text{4}}{\text{3}}\left({\text{ρ}}_{\text{1}}^{\text{3}}{\text{-10}}^{\text{-9}}\right){\text{ μC/m}}^{\text{2}}$

To determine

(c)

The value of ${\text{D}}_{\text{ρ}}$ at different values of $\text{ρ}$.

Answer to Problem 3.15P

${\text{D}}_{\text{ρ}}\text{ at ρ = 0}\text{.8mm}$ is 0

${\text{D}}_{\text{ρ}}\text{ at ρ = 1}\text{.6mm}$ is $\text{3}{\text{.6×10}}^{\text{-6}}{\text{ μ C/m}}^{\text{2}}$

${\text{D}}_{\text{ρ}}\text{ at ρ = 2}\text{.4mm}$ is $\text{3}{\text{.9 × 10}}^{\text{-6}}{\text{ μC/m}}^{\text{2}}$

Explanation of Solution

Given Information:

   $\begin{array}{c}{\rho }_v=0\text{ for }\rho <1\text{mm and }\rho >2\text{mm}\\ {\rho }_v=4\rho \mu {\text{ C/m}}^{\text{3}}\text{ for }1<\rho <2\text{mm}\end{array}$

Concept used:

   ${\text{D}}_{\text{ρ}\left({\text{ρ}}_{\text{1}}\right)}\text{ = }\frac{\text{4}\left({\text{ρ}}_{\text{1}}^{\text{3}}{\text{-10}}^{\text{-9}}\right)}{{\text{3ρ}}_{\text{1}}}{\text{ μ C/m}}^{\text{2}}$

Calculation:

At $\text{ρ = 0}\text{.8mm}$ , no charge is enclosed by a cylindrical Gaussian surface of that radius,

So ${\text{D}}_{\text{ρ}}$ (0.8 mm) = 0

At $\text{ρ = 1}\text{.6 mm}$ ,

   $\begin{array}{c}{D}_{\rho }\left(\text{1}.\text{6 mm}\right)\text{= }\frac{\text{4}\left[{\left(\text{0}\text{.0016}\right)}^{\text{3}}\text{- }\left(\text{0}{\text{.0010}}^{\text{3}}\right)\right]}{\text{3}\left(\text{0}\text{.0016}\right)}\\ =\frac{\text{4}\left[\text{4}{\text{.09 × 10}}^{\text{-9}}{\text{- 1 × 10}}^{\text{-9}}\right]}{\text{0}\text{.0048}}\\ =\frac{\text{16}{\text{.36 × 10}}^{\text{-9}}{\text{- 4 × 10}}^{\text{-9}}}{\text{0}\text{.0048}}\\ =\frac{\text{12}{\text{.36 × 10}}^{\text{-9}}}{\text{0}\text{.0048}}\\ =\text{3}{\text{.6 × 10}}^{\text{-6}}{\text{μ C/m}}^{\text{3}}\end{array}$

At $\text{ρ = 2}\text{.4 mm}$ , we evaluate the charge integral of part a from 0.001 to 0.002 and Gauss's law

is written as

   $\begin{array}{c}{\text{2πρD}}_{\text{ρ}}\text{ = }\frac{\text{8πL}}{\text{3}}\left[{\left(\text{0}\text{.002}\right)}^{\text{2}}\text{-}{\left(\text{0}\text{.001}\right)}^{\text{2}}\right]\text{ μC}\\ \text{= }\frac{\text{7}{\text{.5 × 10}}^{\text{-5}}}{\text{3}}\\ \text{= 2}{\text{.5 × 10}}^{\text{-5}}\\ {\text{D}}_{\text{ρ}}\text{ = }\frac{\text{2}{\text{.5 × 10}}^{\text{-5}}}{\text{2πρ}}\\ \text{= }\frac{\text{2}{\text{.5 × 10}}^{\text{-5}}}{\text{2π × 2}\text{.4}}\\ \text{= 3}{\text{.9 × 10}}^{\text{-6}}{\text{ μC/m}}^{\text{2}}\end{array}$

Conclusion:

${\text{D}}_{\text{ρ}}\text{ at ρ = 0}\text{.8mm}$ is 0

${\text{D}}_{\text{ρ}}\text{ at ρ = 1}\text{.6mm}$ is $\text{3}{\text{.6×10}}^{\text{-6}}{\text{ μ C/m}}^{\text{2}}$

${\text{D}}_{\text{ρ}}\text{ at ρ = 2}\text{.4mm}$ is $\text{3}{\text{.9 × 10}}^{\text{-6}}{\text{ μC/m}}^{\text{2}}$