Chapter 23, Problem 64P

(a)

To determine

The shape of the equipotential surfaces in the region around the charge.

Explanation of Solution

Introduction:

The supplementary and inherent essential property of some subatomic particles such as electron due to which they interact with each other is known as electric charge. The standard unit of electric charge is Coulomb which is symbolized as $\text{C}$ .

The appearance of the equipotential surfaces in the region around the charge is concentric with the middle wire as the geometry of the Geiger tube is cylindrical. An equipotential surface is a surface that is either real or imaginary around a point charge. The electric potential on the equipotential surface is equivalent at each point, and these equipotential surfaces are formed at some distance from the point charge.

Conclusion:

Therefore, the shape of the equipotential surfaces in the region around the charge is concentric with the central or middle wire.

(b)

To determine

The radii of the five surfaces that have potentials equals to $20.0\text{V,40}\text{.0}\text{V,60}\text{.0}\text{V,80}\text{.0}\text{V}$ and $100.0\text{V}$ .

Answer to Problem 64P

The radii of the five surfaces that have potentials equals to $20.0\text{V,40}\text{.0}\text{V,60}\text{.0}\text{V,80}\text{.0}\text{V}$ and $100.0\text{V}$ is tabulated as below and the sketch is shown in Figure 1.

$V\left(\text{V}\right)$	$20.0$	$40.0$	$60.0$	$80.0$	$100.0$
$r\left(\text{m}\right)$	$4.99$	$2.49$	$1.66$	$1.25$	$1.00$

Explanation of Solution

Given:

The potential of first surface is $V_1=20.0\text{V}$

The potential of second surface is $V_2=40.0\text{V}$

The potential of third surface is $V_3=60.0\text{V}$

The potential of fourth surface is $V_4=80.0\text{V}$

The potential of fifth surface is $V_5=100.0\text{V}$

Formula used:

The expression for the relationship between electric potential due to point charge and the electric field of the point charge is given by,

  $\begin{array}{c}{\displaystyle \underset{a}{\overset{b}{\int }}dV}=-{\displaystyle \underset{r_a}{\overset{r_b}{\int }}\overrightarrow{E}}\cdot dr\\ =-kQ{\displaystyle \underset{r_a}{\overset{r_b}{\int }}\left(\frac{1}{r^2}\right)dr}\\ =V_b-V_a=kQ\left(\frac{1}{r_b}-\frac{1}{r_a}\right)\end{array}$

Calculation:

Suppose the potential to be zero at $r_a=\infty$ ,

  $\begin{array}{c}{V}_b-0=kQ\left(\frac{1}{r_b}\right)\\ r=\frac{kQ}{V}\end{array}$

The radius for $V=20.0\text{V}$ is calculated as,

  $\begin{array}{c}r=\frac{kQ}{V}\\ =\frac{\left(8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(1.11\times {10}^{-8}\text{C}\right)}{20.0\text{V}}\\ =4.99\text{m}\end{array}$

The radius for $V=40.0\text{V}$ is calculated as,

  $\begin{array}{c}r=\frac{kQ}{V}\\ =\frac{\left(8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(1.11\times {10}^{-8}\text{C}\right)}{40.0\text{V}}\\ =2.49\text{m}\end{array}$

The radius for $V=60.0\text{V}$ is calculated as,

  $\begin{array}{c}r=\frac{kQ}{V}\\ =\frac{\left(8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(1.11\times {10}^{-8}\text{C}\right)}{60.0\text{V}}\\ =1.66\text{m}\end{array}$

The radius for $V=80.0\text{V}$ is calculated as,

  $\begin{array}{c}r=\frac{kQ}{V}\\ =\frac{\left(8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(1.11\times {10}^{-8}\text{C}\right)}{80.0\text{V}}\\ =1.25\text{m}\end{array}$

The radius for $V=100.0\text{V}$ is calculated as,

  $\begin{array}{c}r=\frac{kQ}{V}\\ =\frac{\left(8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(1.11\times {10}^{-8}\text{C}\right)}{100.0\text{V}}\\ =1.00\text{m}\end{array}$

The sketch is shown below.

  

Figure 1

Conclusion:

Therefore, the radii of the five surfaces that have potentials equals to $20.0\text{V,40}\text{.0}\text{V,60}\text{.0}\text{V,80}\text{.0}\text{V}$ and $100.0\text{V}$ is tabulated as below and sketch is shown in Figure 1.

$V\left(\text{V}\right)$	$20.0$	$40.0$	$60.0$	$80.0$	$100.0$
$r\left(\text{m}\right)$	$4.99$	$2.49$	$1.66$	$1.25$	$1.00$

(c)

To determine

The spacing between the equipotential surfaces.

Explanation of Solution

Introduction:

The surface where the magnitude of gravitational potential is the same at all points is known as equipotential, and it coincides with the directions of the gravitational force. A liquid surface is in equilibrium is the best example to understand the concept of the equipotential surface.

The spacing between the equipotential surfaces is closest together but are not equally spaced, it means the electric field becomes stronger. If the field is not performing any work on the particle as it travels from one position to another then the direction of the field’s force must be normal to the direction of the movement of the particle.

Conclusion:

Therefore, the spacing between the equipotential surfaces is closest together

(d)

To determine

The electric field strength between $40.0\text{V}$ and $60.0\text{V}$ equipotential surfaces.

Answer to Problem 64P

The electric field strength between $40.0\text{V}$ and $60.0\text{V}$ equipotential surfaces is $23\text{V}/\text{m}$

Explanation of Solution

Given:

The potential of equipotential surface has $40\text{V}$ and $60\text{V}$ .

Formula used:

The expression for the average value of the magnitude of the electric field is given as,

  $E=-\frac{\Delta V}{\Delta r}$

The expression for the exact value of the electric field strength is given as,

  $E=\frac{kQ}{r^2}$

Calculation:

The average value of the magnitude of the electric field is calculated as,

  $\begin{array}{c}E=-\frac{\Delta V}{\Delta r}\\ =-\frac{\left(40\text{V}\right)-\left(60\text{V}\right)}{\left(2.4\text{m}\right)-\left(1.7\text{m}\right)}\\ =29\text{V}/\text{m}\end{array}$

The exact value of the electric field strength is calculated as,

  $\begin{array}{c}E=\frac{kQ}{r^2}\\ =\frac{\left(8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(1.11\times {10}^{-8}\text{C}\right)}{{\left(\frac{1.66\text{m}+2.49\text{m}}{2}\right)}^2}\\ =23\text{V}/\text{m}\end{array}$

Conclusion:

Therefore, the electric field strength between $40.0\text{V}$ and $60.0\text{V}$ equipotential surfaces is $23\text{V}/\text{m}$