Chapter 16, Problem 66QAP

To determine

(a)

The electric flux out of each face.

Answer to Problem 66QAP

The electric flux out of each face is, ${\varphi }_I=0\text{Vm}$

  ${\varphi }_{II}=48\text{Vm}$

  ${\varphi }_{III}=48\text{Vm}$

  ${\varphi }_{IV}=48\text{Vm}$

  ${\varphi }_V=0\text{Vm}$

Explanation of Solution

Given:

Height of prism, $h=40\text{cm}$

Depth of prism, $d=30\text{cm}$

Length of prism, $l=80\text{cm}$

Electric field, $E=500\text{N/C}$

Formula used:

The electric flux is given by,
  $\varphi =EA\cos \theta$

Where,
  $\varphi$ =Electric flux
  $E$ = Electric field
  $A$ =Area of prism

Calculation:

The electric flux is given by,
  $\varphi =EA\cos \theta$

For surface I: $\theta =90°\text{C}$

Thus,
  $\begin{array}{l}{\varphi }_I=EA\cos 90\\ {\varphi }_I=0\text{Vm}\end{array}$

For surface II: $\theta =0°\text{C}$

Thus,
  $\begin{array}{l}{\varphi }_{II}=EA\cos 0\\ {\varphi }_{II}=500\times 0.4\times 0.3\times 0.8\times 1\text{Vm}\\ {\varphi }_{II}=48\text{Vm}\end{array}$

For surface III: $\theta =0°\text{C}$

Thus,
  $\begin{array}{l}{\varphi }_{III}=EA\cos 0\\ {\varphi }_{III}=500\times 0.4\times 0.3\times 0.8\times 1\text{Vm}\\ {\varphi }_{III}=48\text{Vm}\end{array}$

For surface IV: $\theta =0°\text{C}$

Thus,
  $\begin{array}{l}{\varphi }_{IV}=EA\cos 0\\ {\varphi }_{IV}=500\times 0.4\times 0.3\times 0.8\times 1\text{Vm}\\ {\varphi }_{IV}=48\text{Vm}\end{array}$

For surface V: $\theta =90°\text{C}$

Thus,
  $\begin{array}{l}{\varphi }_V=EA\cos 90\\ {\varphi }_V=0\text{Vm}\end{array}$

To determine

(b)

The net electric flux.

Answer to Problem 66QAP

The net electric flux is, $\Phi =164\text{Vm}$

Explanation of Solution

Height of prism, $h=40\text{cm}$

Depth of prism, $d=30\text{cm}$

Length of prism, $l=80\text{cm}$

Electric field, $E=500\text{N/C}$

Formula used:

The electric flux is given by,
  $\varphi =EA\cos \theta$

Where,
  $\varphi$ =Electric flux
  $E$ = Electric field
  $A$ =Area of prism

Calculation:

The net electric flux is,
  $\begin{array}{l}\Phi ={\varphi }_I+{\varphi }_{II}+{\varphi }_{III}+{\varphi }_{IV}+{\varphi }_V\\ \Phi =0+48+48+48+0\\ \Phi =164\text{Vm}\end{array}$

To determine

(c)

The electric field the prism enclosed.

Answer to Problem 66QAP

The electric field enclosed by the prism is, $E_{net}=-199500\text{N/C}$

Explanation of Solution

Given:

Height of prism, $h=40\text{cm}$

Depth of prism, $d=30\text{cm}$

Length of prism, $l=80\text{cm}$

Electric field, $E=500\text{N/C}$

Radius of solid sphere, $R$

Charge, $Q=-2.00\mu \text{C}$

Formula used:

The electric flux is given by,
  $E=\frac{Q}{A{\epsilon }_0}$

Where,
  $E$ =Electric flux
  $Q$ = Charge
A=Surface area of prism

Calculation:

The electric flux is given by,
  $E_Q=\frac{Q}{A{\epsilon }_0}$

Area of the prism is,
  $\begin{array}{l}A=3\left(l\times h\right)+\left(h\times d\right)\\ A=3\left(0.8\times 0.4\right)+\left(0.4\times 0.3\right)\\ A=1.08{\text{m}}^2\end{array}$

Thus,
  $\begin{array}{l}{E}_Q=\frac{-2.00\times {10}^{-6}}{1.08\times 8.85\times {10}^{-12}}\\ E_Q=-2.09\times {10}^5\text{N/C}\end{array}$

So, electric field of the prism is,

$\begin{array}{l}{E}_{net}=E+E_Q\\ E_{net}=500-2.00\times {10}^5\\ E_{net}=-199500\text{N/C}\end{array}$