Chapter 4, Problem 48P

(a)

To determine

The energy expended in transferring $10\text{nC}$ charge from $A\left(5,30°,0°\right)$ to $B\left(5,90°,0°\right)$.

Explanation of Solution

Given:

The electric field intensity $\left(E\right)$ is $20r\sin \theta a_r+10r\cos \theta a_{\theta }\text{V}/\text{m}$.

The first point $\left(A\right)$ is $\left(5,30°,0°\right)$.

The second point $\left(B\right)$ is $\left(5,90°,0°\right)$.

The magnitude of the charge $\left(Q\right)$ is $10\text{nC}$.

Calculation:

The energy expended to transfer the charge is equal to the work done to move the charge.

Calculate the work done $\left(W\right)$ using the relation.

  $W=-Q{\displaystyle {\int }_{L}E\cdot dl}$

  $\begin{array}{l}W=-\left(10\text{nC}\right){\displaystyle \underset{A\left(5,30°,0°\right)}{\overset{B\left(5,90°,0°\right)}{\int }}\left(20r\sin \theta a_r+10r\cos \theta a_{\theta }\text{V}/\text{m}\right)}\cdot \left(dr{a}_r+rd\theta a_{\theta }+r\sin \theta d\varphi a_{\varphi }\right)\\ W=-\left(10\text{nC}\right){{\displaystyle \underset{\theta =30}{\overset{90}{\int }}\left[10r^2\cos \theta d\theta \right]}}_{r=5,\varphi =0°}\\ W=-\left(10\text{nC}\right)\left(10\times 5^2\right){\left[\sin \theta \right]}_{30°}^{90°}\\ W=-1250\text{nJ}\end{array}$

Thus, the energy expended to transfer $10\text{nC}$ charge from $A\left(5,30°,0°\right)$ to $B\left(5,90°,0°\right)$ is $\underset{\_}{-1250\text{nJ}}$.

(b)

To determine

The energy expended in transferring $10\text{nC}$ charge from $A\left(5,30°,0°\right)$ to $C\left(10,30°,0°\right)$.

Explanation of Solution

Given:

The electric field intensity $\left(E\right)$ is $12\rho z\cos \varphi a_{\rho }-6\rho z{a}_{\varphi }+6{\rho }^2\cos \varphi a_z$.

The first point $\left(A\right)$ is $\left(5,30°,0°\right)$

The second point $\left(C\right)$ is $\left(10,30°,0°\right)$.

The magnitude of the charge $\left(Q\right)$ is $10\text{nC}$.

Calculation:

Calculate the work done $\left(W\right)$ using the relation.

  $W=-Q{\displaystyle {\int }_{L}E\cdot dl}$

  $\begin{array}{l}W=-\left(10\text{nC}\right){\displaystyle \underset{A\left(5,30°,0°\right)}{\overset{B\left(10,30°,0°\right)}{\int }}\left(20r\sin \theta a_r+10r\cos \theta a_{\theta }\text{V}/\text{m}\right)}\cdot \left(dr{a}_r+rd\theta a_{\theta }+r\sin \theta d\varphi a_{\varphi }\right)\\ W=-\left(10\text{nC}\right){{\displaystyle \underset{r=5}{\overset{10}{\int }}\left[20r\sin \theta dr\right]}}_{\theta =30°,\varphi =0°}\\ W=-\left(10\text{nC}\right)\left(20\times \sin \left(30°\right)\right){\left[\frac{r^2}{2}\right]}_5^{10}\\ W=-\left(10\text{nC}\right)\left(20\times \sin \left(30°\right)\right)\left[\frac{{10}^2}{2}-\frac{5^2}{2}\right]\end{array}$

  $W=-3750\text{nJ}$

Thus, the energy expended to transfer $10\text{nC}$ charge from $A\left(5,30°,0°\right)$ to $C\left(10,30°,0°\right)$ is $\underset{\_}{-3750\text{nJ}}$.

(c)

To determine

The energy expended in transferring $10\text{nC}$ charge from $A\left(5,30°,0°\right)$ to $D\left(5,30°,60°\right)$.

Explanation of Solution

Given:

The electric field intensity $\left(E\right)$ is $12\rho z\cos \varphi a_{\rho }-6\rho z{a}_{\varphi }+6{\rho }^2\cos \varphi a_z$.

The first point $\left(A\right)$ is $\left(5,30°,0°\right)$ and the second point $\left(D\right)$ is $\left(5,30°,60°\right)$.

Calculation:

Calculate the work done $\left(W\right)$ using the relation.

  $W=-Q{\displaystyle {\int }_{L}E\cdot dl}$

  $\begin{array}{l}W=-\left(10\text{nC}\right){\displaystyle \underset{A\left(5,30°,0°\right)}{\overset{B\left(5,30°,60°\right)}{\int }}\left(20r\sin \theta a_r+10r\cos \theta a_{\theta }\text{V}/\text{m}\right)}\cdot \left(dr{a}_r+rd\theta a_{\theta }+r\sin \theta d\varphi a_{\varphi }\right)\\ W=-\left(10\text{nC}\right){{\displaystyle \underset{\varphi =0°}{\overset{60°}{\int }}\left[0d\varphi \right]}}_{r=5,\theta =30°}\\ W=0\text{nJ}\end{array}$

Thus, the energy expended to transfer $10\text{nC}$ charge from $A\left(5,30°,0°\right)$ to $D\left(5,30°,60°\right)$ is $\underset{\_}{0\text{nJ}}$.

(d)

To determine

The energy expended in transferring $10\text{nC}$ charge from $A\left(5,30°,0°\right)$ to $E\left(10,90°,60°\right)$.

Explanation of Solution

Given:

The electric field intensity $\left(E\right)$ is $12\rho z\cos \varphi a_{\rho }-6\rho z{a}_{\varphi }+6{\rho }^2\cos \varphi a_z$.

The first point $\left(A\right)$ is $\left(5,30°,0°\right)$

The second point $\left(E\right)$ is $\left(10,90°,60°\right)$.

The magnitude of the charge $\left(Q\right)$ is $10\text{nC}$.

Calculation:

Since the electrostatic force is conservative, the path of integration is immaterial.

Suppose the charge follow the path to from point $A\left(5,30°,0°\right)$ to $B\left(5,90°,0°\right)$ and then $B\left(5,90°,0°\right)$ to $B^{\prime }\left(10,90°,0°\right)$ and then $B^{\prime }\left(10,90°,0°\right)$ to $E\left(10,90°,60°\right)$.

Calculate the work done $\left(W\right)$ by using the relation.

  $W=-Q{\displaystyle {\int }_{L}E\cdot dl}$

  $\begin{array}{l}W=-\left(10\text{nC}\right){\displaystyle \underset{A\left(5,30°,0°\right)}{\overset{E\left(10,90°,60°\right)}{\int }}\left(20r\sin \theta a_r+10r\cos \theta a_{\theta }\text{V}/\text{m}\right)}\cdot \left(dr{a}_r+rd\theta a_{\theta }+r\sin \theta d\varphi a_{\varphi }\right)\\ W=-\left(10\text{nC}\right){\displaystyle \underset{\theta =30°}{\overset{90°}{\int }}{\displaystyle \underset{r=5}{\overset{10}{\int }}{\displaystyle \underset{\varphi =0°}{\overset{60°}{\int }}\left(20r\sin \theta a_r+10r\cos \theta a_{\theta }\text{V}/\text{m}\right)\cdot \left(dr{a}_r+rd\theta a_{\theta }+r\sin \theta d\varphi a_{\varphi }\right)}}}\\ W=-\left(10\text{nC}\right)\left[\left[{\displaystyle \underset{\theta =30°}{\overset{90°}{\int }}{\left[10r^2\cos \theta d\theta \right]}_{r=5,\varphi =0°}}\right]+\left[{\displaystyle \underset{r=5}{\overset{r=10}{\int }}{\left[20r\sin \theta dr\right]}_{\theta =90°,\varphi =0°}}\right]+\left[{\displaystyle \underset{\varphi =0°}{\overset{60°}{\int }}0d\varphi }\right]\right]\\ W=-\left(10\text{nC}\right)\left[\left(10\times 5^2\right){\left[\sin \theta \right]}_{30°}^{90°}+20\sin \left(90°\right){\left[\frac{r^2}{2}\right]}_5^{10}+0\right]\end{array}$

  $\begin{array}{l}W=-\left(10\text{nC}\right)\left[\left(10\times 5^2\right)\left(1-0.5\right)+20\sin \left(90°\right)\left[\frac{{10}^2}{2}-\frac{5^2}{2}\right]\right]\\ W=-8750\text{nJ}\end{array}$

Thus, the energy expended to transfer $10\text{nC}$ charge from $A\left(5,30°,0°\right)$ to $E\left(10,90°,60°\right)$ is $\underset{\_}{-8750\text{nJ}}$.