Chapter 6, Problem 6.39P

To determine

(a)

The voltage $V\left(\varphi \right)$.

Answer to Problem 6.39P

The voltage $V\left(\varphi \right)$ is $\left[\left(-2\times {10}^4\right)\varphi +\left(3.78\times {10}^3\right)\right]\text{V}$.

Explanation of Solution

Calculation:

The general solution of Laplace's equation is written as,

   $V=C_1\varphi +C_2$ ...... (1)

Here,

   $\varphi$ is the spherical coordinate.

   $C_1$ and $C_2$ are constant values.

Substitute $20\text{V}$ for $V$ and $0.188$ for $\varphi$ in equation (1).

   $20\text{V}=C_1\left(0.188\right)+C_2$ ...... (2)

Equation (2) is simplified as,

   $20\text{V}-C_1\left(0.188\right)+C_2=0$ ...... (3)

Substitute $200\text{V}$ for $V$ and $0.179$ for $\varphi$ in equation (1).

   $200\text{V}=C_1\left(0.179\right)+C_2$ ...... (4)

Equation (4) is simplified as,

   $200\text{V}-C_1\left(0.179\right)+C_2=0$ ...... (5)

Subtracting equation (5) from equation (3) as,

   $\left(20\text{V}-C_1\left(0.188\right)+C_2=0\right)-\left(200\text{V}-C_1\left(0.179\right)+C_2=0\right)$ ...... (6)

By solving equation (6) the values of $C_1$ and $C_2$ is,

   $\begin{array}{l}{C}_1=-2\times {10}^4\\ C_2=3.78\times {10}^3\end{array}$

Substitute $-2\times {10}^4$ for $C_1$ and $3.78\times {10}^3$ for $C_2$ in equation (1).

   $V=\left[\left(-2\times {10}^4\right)\varphi +\left(3.78\times {10}^3\right)\right]\text{V}$ ...... (7)

Conclusion:

Therefore, the voltage $V\left(\varphi \right)$ is $\left[\left(-2\times {10}^4\right)\varphi +\left(3.78\times {10}^3\right)\right]\text{V}$.

To determine

(b)

The electric field $\text{E}\left(\rho \right)$.

Answer to Problem 6.39P

The electric field intensity $\text{E}\left(\rho \right)$ is $\frac{2\times {10}^4}{\rho }\text{V/m}$.

Explanation of Solution

Calculation:

The electric field $\text{E}\left(\rho \right)$ is written as,

   $\text{E}\left(\rho \right)=-\frac{1}{\rho }\frac{dV}{d\varphi }$ ...... (8)

Here,

   $\rho$ is the cylindrical coordinate.

Differentiate the equation (7) with respect to $\varphi$ as,

   $\frac{dV}{d\varphi }=-2\times {10}^4\text{V/m}$

Substitute $-2\times {10}^4\text{V/m}$ for $\frac{dV}{d\varphi }$ in equation (8).

   $\begin{array}{c}\text{E}\left(\rho \right)=-\frac{1}{\rho }\left(-2\times {10}^4\text{V/m}\right){\text{a}}_{\varphi }\\ =\frac{2\times {10}^4}{\rho }{\text{a}}_{\varphi }\text{V/m}\end{array}$

Conclusion:

Therefore, the electric field intensity $\text{E}\left(\rho \right)$ is $\frac{2\times {10}^4}{\rho }{\text{a}}_{\varphi }\text{V/m}$.

To determine

(c)

The electric field displacement $\text{D}\left(\rho \right)$.

Answer to Problem 6.39P

The electric field displacement $\text{D}\left(\rho \right)$ is $\frac{17.7\times {10}^{-8}}{\rho }{\text{a}}_{\varphi }{\text{C/m}}^2$.

Explanation of Solution

Calculation:

The electric field displacement $\text{D}\left(\rho \right)$ is written as,

   $\text{D}\left(\rho \right)={\epsilon }_0\text{E}\left(\rho \right)$ ...... (9)

Here,

   ${\epsilon }_0$ is the absolute permittivity.

Substitute $\frac{2\times {10}^4}{\rho }{\text{a}}_{\varphi }\text{V/m}$ for $\text{E}\left(\rho \right)$ and $8.85\times {10}^{-12}\text{F/m}$ for ${\epsilon }_0$ in equation (9).

   $\begin{array}{c}\text{D}\left(\rho \right)=\left(8.85\times {10}^{-12}\text{F/m}\right)\left(\frac{2\times {10}^4}{\rho }{\text{a}}_{\varphi }\text{V/m}\right)\\ =\frac{17.7\times {10}^{-8}}{\rho }{\text{a}}_{\varphi }{\text{C/m}}^2\end{array}$

Conclusion:

Therefore, the electric field displacement $\text{D}\left(\rho \right)$ is $\frac{17.7\times {10}^{-8}}{\rho }{\text{a}}_{\varphi }{\text{C/m}}^2$.

To determine

(d)

The charge surface density ${\rho }_{\text{s}}$.

Answer to Problem 6.39P

The charge surface density is $\frac{17.7\times {10}^{-8}}{\rho }{\text{C/m}}^2$.

Explanation of Solution

Calculation:

The charge surface density ${\rho }_{\text{s}}$ is written as,

   ${\rho }_{\text{s}}=\text{D}\cdot \text{n}$ ...... (10)

Here,

   $\text{n}$ is the direction of electrical field.

Substitute $\frac{17.7\times {10}^{-8}}{\rho }{\text{a}}_{\varphi }{\text{C/m}}^2$ for $\text{D}$ and $a_{\varphi }$ for $\text{n}$ in equation (10).

   $\begin{array}{c}{\rho }_{\text{s}}=\left(\frac{17.7\times {10}^{-8}}{\rho }{\text{a}}_{\varphi }{\text{c/m}}^2\right)\cdot {\text{a}}_{\varphi }\\ =\frac{17.7\times {10}^{-8}}{\rho }{\text{C/m}}^2\end{array}$

Conclusion:

Therefore, the charge surface density is $\frac{17.7\times {10}^{-8}}{\rho }{\text{C/m}}^2$.

To determine

(e)

The charge $Q$ on the upper surface of the lower plane.

Answer to Problem 6.39P

The charge $Q$ on the upper surface of the lower plane is $8.47\times {10}^{-8}\text{C}$.

Explanation of Solution

Calculation:

The charge $Q$ on the upper surface of the lower plane is written as,

   $Q={\displaystyle {\int }_{z_0}^{z_{\text{n}}}{\displaystyle {\int }_{{\rho }_0}^{{\rho }_{\text{n}}}{\rho }_{\text{s}}}}d\rho dz$ ...... (11)

Here,

   $z_0$ is the initial value of cylindrical coordinate $z$.

   $z_{\text{n}}$ is the final value of cylindrical coordinate $z$.

   ${\rho }_0$ is the initial value of cylindrical coordinate $\rho$.

   ${\rho }_{\text{n}}$ is the final value of cylindrical coordinate $\rho$.

Substitute $0$ for $z_0$ , $0.1$ for $z_{\text{n}}$ , $0.001$ for ${\rho }_0$ , $0.120$ for ${\rho }_{\text{n}}$ , and $\frac{17.7\times {10}^{-8}}{\rho }{\text{C/m}}^2$ for ${\rho }_{\text{s}}$ in equation (11).

   $\begin{array}{c}Q={\displaystyle {\int }_0^{0.1}{\displaystyle {\int }_{0.001}^{0.120}\left(\frac{17.7\times {10}^{-8}}{\rho }{\text{C/m}}^2\right)}}d\rho dz\\ =8.47\times {10}^{-8}\text{C}\end{array}$

Conclusion:

Therefore, the charge $Q$ on the upper surface of the lower plane is $8.47\times {10}^{-8}\text{C}$.

To determine

(f)

The voltage $V\left(\varphi \right)$ , the electric field $E\left(\varphi \right)$ , the electric field displacement $\text{D}\left(\rho \right)$ for upper plane $\varphi =0.188-2\pi$ and the charge surface density ${\rho }_{\text{s}}$ , the charge $Q$ on the lower surface of the lower plane.

Answer to Problem 6.39P

The voltage $V\left(\varphi \right)$ is $\left(28.7\varphi +194.9\right)\text{V}$ , the electric field $E\left(\varphi \right)$ is $-\frac{28.7}{\rho }{\text{a}}_{\varphi }\text{V/m}$ , the electric field displacement $\text{D}\left(\rho \right)$ is $-\frac{253.995\times {10}^{-12}}{\rho }{\text{a}}_{\varphi }{\text{C/m}}^2$ , the charge surface density ${\rho }_{\text{s}}$ is $\frac{253.995\times {10}^{-12}}{\rho }{\text{C/m}}^2$ , and the charge $Q$ is $122\times {10}^{-12}\text{C}$.

Explanation of Solution

Calculation:

The general solution of Laplace's equation is written as,

   $V=C_1{}^{\prime }\varphi +C_2{}^{\prime }$ ...... (12)

Substitute $20\text{V}$ for $V$ and $0.188-2\pi$ for $\varphi$ in equation (12).

   $20\text{V}=C_1{}^{\prime }\left(0.188-2\pi \right)+C_2{}^{\prime }$ ...... (13)

Equation (13) is simplified as,

   $20\text{V}-C_1{}^{\prime }\left(0.188-2\pi \right)+C_2{}^{\prime }=0$ ...... (14)

Substitute $200\text{V}$ for $V$ and $0.179$ for $\varphi$ in equation (12).

   $200\text{V}=C_1{}^{\prime }\left(0.179\right)+C_2{}^{\prime }$ ...... (15)

Equation (15) is simplified as,

   $200\text{V}-C_1{}^{\prime }\left(0.179\right)+C_2{}^{\prime }=0$ ...... (16)

Subtracting equation (16) from equation (14) as,

   $\left(20\text{V}-C_1{}^{\prime }\left(0.188-2\pi \right)+C_2{}^{\prime }=0\right)-\left(200\text{V}-C_1{}^{\prime }\left(0.179\right)+C_2{}^{\prime }=0\right)$ ...... (17)

By solving equation (17) the values of $C_1{}^{\prime }$ and $C_2{}^{\prime }$ is,

   $\begin{array}{l}{C}_1{}^{\prime }=28.7\\ C_2{}^{\prime }=194.9\end{array}$

Substitute $28.7$ for $C_1{}^{\prime }$ and $194.9$ for $C_2{}^{\prime }$ in equation (12).

   $V=\left[28.7\varphi +194.9\right]\text{V}$ ...... (18)

The electric field $E\left(\rho \right)$ is written as,

   $\text{E}\left(\rho \right)=-\frac{1}{\rho }\frac{dV}{d\varphi }$ ...... (19)

Differentiate the equation (18) with respect to $\varphi$ as,

   $\frac{dV}{d\varphi }=28.7\text{V/m}$

Substitute $28.7\text{V/m}$ for $\frac{dV}{d\varphi }$ in equation (19).

   $\text{E}\left(\rho \right)=-\frac{28.7}{\rho }{\text{a}}_{\varphi }\text{V/m}$

The electric field displacement $\text{D}\left(\rho \right)$ is written as,

   $\text{D}\left(\rho \right)={\epsilon }_0\text{E}\left(\rho \right)$ ...... (20)

Substitute $-\frac{28.7}{\rho }{\text{a}}_{\varphi }\text{V/m}$ for $\text{E}\left(\rho \right)$ and $8.85\times {10}^{-12}\text{F/m}$ for ${\epsilon }_0$ in equation (20).

   $\begin{array}{c}\text{D}\left(\rho \right)=\left(8.85\times {10}^{-12}\text{F/m}\right)\left(-\frac{28.7}{\rho }{\text{a}}_{\varphi }\text{V/m}\right)\\ =-\frac{253.995\times {10}^{-12}}{\rho }{\text{a}}_{\varphi }{\text{C/m}}^2\end{array}$

The charge surface density ${\rho }_{\text{s}}$ is written as,

   ${\rho }_{\text{s}}=\text{D}\cdot \text{n}$ ...... (21)

Substitute $-\frac{253.995\times {10}^{-12}}{\rho }{\text{a}}_{\varphi }{\text{C/m}}^2$ for $\text{D}$ and $-a_{\varphi }$ for $\text{n}$ in equation (21).

   $\begin{array}{c}{\rho }_{\text{s}}=\left(-\frac{253.995\times {10}^{-12}}{\rho }{\text{a}}_{\varphi }{\text{C/m}}^2\right)\cdot \left(-{\text{a}}_{\varphi }\right)\\ =\frac{253.995\times {10}^{-12}}{\rho }{\text{C/m}}^2\end{array}$

The charge $Q$ on the lower surface of the lower plane is written as,

   $Q={\displaystyle {\int }_{z_0}^{z_{\text{n}}}{\displaystyle {\int }_{{\rho }_0}^{{\rho }_{\text{n}}}{\rho }_{\text{s}}}}d\rho dz$ ...... (22)

Substitute $0$ for $z_0$ , $0.1$ for $z_{\text{n}}$ , $0.001$ for ${\rho }_0$ , $0.120$ for ${\rho }_{\text{n}}$ , and $\frac{253.995\times {10}^{-12}}{\rho }{\text{C/m}}^2$ for ${\rho }_{\text{s}}$ in equation (22).

   $\begin{array}{c}Q={\displaystyle {\int }_0^{0.1}{\displaystyle {\int }_{0.001}^{0.120}\left(\frac{253.995\times {10}^{-12}}{\rho }{\text{C/m}}^2\right)}}d\rho dz\\ =122\times {10}^{-12}\text{C}\end{array}$

Conclusion:

Therefore, the voltage $V\left(\varphi \right)$ is $\left(28.7\varphi +194.9\right)\text{V}$ , the electric field $E\left(\varphi \right)$ is $-\frac{28.7}{\rho }{\text{a}}_{\varphi }\text{V/m}$ , the electric field displacement $\text{D}\left(\rho \right)$ is $-\frac{253.995\times {10}^{-12}}{\rho }{\text{a}}_{\varphi }{\text{C/m}}^2$ , the charge surface density ${\rho }_{\text{s}}$ is $\frac{253.995\times {10}^{-12}}{\rho }{\text{C/m}}^2$ , and the charge $Q$ is $122\times {10}^{-12}\text{C}$.

(g)

To determine

The total charge on the lower plane and the capacitance between the planes.

Answer to Problem 6.39P

The total charge on the lower plane is $84.8\times {10}^{-9}\text{C}$ and the capacitance between the planes is $471\times {10}^{-12}\text{F}$.

Explanation of Solution

Calculation:

The total charge $Q_{\text{net}}$ is written as,

   $Q_{\text{net}}=Q_t+Q_b$ ...... (23)

Here,

   $Q_t$ is the charge on the upper surface of the lower plane.

   $Q_b$ is the charge on the lower surface of the lower plane.

Substitute $122\times {10}^{-12}\text{C}$ for $Q_b$ and $8.47\times {10}^{-8}\text{C}$ for $Q_t$ in equation (23).

   $\begin{array}{c}{Q}_{\text{net}}=\left(8.47\times {10}^{-8}\text{C}\right)+\left(122\times {10}^{-12}\text{C}\right)\\ =84.8\times {10}^{-9}\text{C}\end{array}$

The potential difference $\nabla V$ is written as,

   $\nabla V=V_2-V_1$ ...... (24)

Substitute $200\text{V}$ for $V_2$ and $20\text{V}$ for $V_1$ in equation (24).

   $\begin{array}{c}\nabla V=200\text{V}-20\text{V}\\ \text{=180}\text{V}\end{array}$

The capacitance $C$ is written as,

   $C=\frac{Q_{\text{net}}}{\Delta V}$ ...... (25)

Substitute $180\text{V}$ for $\nabla V$ and $84.8\times {10}^{-9}\text{C}$ for $Q_{\text{net}}$ in equation (25).

   $\begin{array}{c}C=\frac{84.8\times {10}^{-9}\text{C}}{180\text{V}}\\ =471\times {10}^{-12}\text{F}\end{array}$

Conclusion:

Therefore, the total charge on the lower plane is $84.8\times {10}^{-9}\text{C}$ and the capacitance between the planes is $471\times {10}^{-12}\text{F}$.