Chapter 20, Problem 7P

(a) Find the potential at a distance of 1.00 cm from a proton. (b) What is the potential difference between two points that are 1.00 cm and 2.00 cm from a proton? (c) Repeat parts (a) and (b) for an electron.

To determine

The potential difference at a distance from proton.

Answer to Problem 7P

The potential difference at a distance from proton is $\underset{\_}{1.44\times {10}^{-7}\text{V}}$.

Explanation of Solution

Write the equation for the potential difference.

  $V_1=\frac{kq}{r_1}$        (I)

Here, $V_1$ is the potential difference for first distance, $k$ is Coulomb’s constant, $q$ is charge and $r_1$ is distance.

Conclusion:

Substitute $8.99\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $1.6\times {10}^{-19}\text{C}$ for $q$, $1.00\text{cm}$ for $r_1$ in Equation (I) to find $V_1$.

  $\begin{array}{c}{V}_1=\frac{\left(8.99\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(1.6\times {10}^{-19}\text{C}\right)}{\left[\left(1.00\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}\\ =1.44\times {10}^{-7}\text{V}\end{array}$

Thus, the potential difference at a distance from proton is $\underset{\_}{1.44\times {10}^{-7}\text{V}}$.

To determine

The potential difference between the points from proton.

Answer to Problem 7P

The potential difference between the points from proton is $\underset{\_}{-7.19\times {10}^{-8}\text{V}}$.

Explanation of Solution

Write the expression for the potential difference between the points from proton.

$\left(\Delta V\right)=V_2-V_1$        (II)

Write the equation for the potential difference at a distance 1.00cm.

  $V_1=\frac{kq}{r_1}$        (III)

Here, $V_1$ is the potential difference for first distance, $k$ is Coulomb’s constant, $q$ is charge and $r_1$ is distance.

Write the equation for the potential difference at a distance 2.00cm.

  $V_2=\frac{kq}{r_2}$        (IV)

Here, $V_2$ is the potential difference for second distance, $k$ is Coulomb’s constant, $q$ is charge and $r_2$ is distance.

Rewrite the expression (II) by using (III) and (IV).

$\left(\Delta V\right)=\left(\frac{kq}{r_2}\right)-\left(\frac{kq}{r_1}\right)$        (V)

Conclusion:

Substitute $8.99\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $1.6\times {10}^{-19}\text{C}$ for $q$, $1.00\text{cm}$ for $r_1$, $2.00\text{cm}$ for $r_2$ in Equation (V) to find $\left(\Delta V\right)$.

$\begin{array}{c}\left(\Delta V\right)=\left[\frac{\left(8.99\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(1.6\times {10}^{-19}\text{C}\right)}{\left[\left(2.00\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}\right]-\left[\frac{\left(8.99\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(1.6\times {10}^{-19}\text{C}\right)}{\left[\left(1.00\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}\right]\\ =-7.19\times {10}^{-8}\text{V}\end{array}$

Thus, the potential difference between the points from proton is $\underset{\_}{-7.19\times {10}^{-8}\text{V}}$.

To determine

The potential difference at a distance from electron.

Answer to Problem 7P

The potential difference at a distance from electron is $\underset{\_}{-1.44\times {10}^{-7}\text{V}}$.

Explanation of Solution

Write the equation for the potential difference.

  $V_1=\frac{kq}{r_1}$        (I)

Here, $V_1$ is the potential difference for first distance, $k$ is Coulomb’s constant, $q$ is charge and $r_1$ is distance.

Conclusion:

Substitute $8.99\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $-1.6\times {10}^{-19}\text{C}$ for $q$, $1.00\text{cm}$ for $r_1$ in Equation (I) to find $V_1$.

  $\begin{array}{c}{V}_1=\frac{\left(8.99\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(-1.6\times {10}^{-19}\text{C}\right)}{\left[\left(1.00\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}\\ =-1.44\times {10}^{-7}\text{V}\end{array}$

Thus, the potential difference at a distance from electron is $\underset{\_}{-1.44\times {10}^{-7}\text{V}}$.

To determine

The potential difference between the points from electron.

Answer to Problem 7P

The potential difference between the points from electron is $\underset{\_}{7.19\times {10}^{-8}\text{V}}$.

Explanation of Solution

Write the expression for the potential difference between the points from electron.

$\left(\Delta V\right)=V_2-V_1$        (II)

Write the equation for the potential difference at a distance 1.00cm.

  $V_1=\frac{kq}{r_1}$        (III)

Here, $V_1$ is the potential difference for first distance, $k$ is Coulomb’s constant, $q$ is charge and $r_1$ is distance.

Write the equation for the potential difference at a distance 2.00cm.

  $V_2=\frac{kq}{r_2}$        (IV)

Here, $V_2$ is the potential difference for second distance, $k$ is Coulomb’s constant, $q$ is charge and $r_2$ is distance.

Rewrite the expression (II) by using (III) and (IV).

$\left(\Delta V\right)=\left(\frac{kq}{r_2}\right)-\left(\frac{kq}{r_1}\right)$        (V)

Conclusion:

Substitute $8.99\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $-1.6\times {10}^{-19}\text{C}$ for $q$, $1.00\text{cm}$ for $r_1$, $2.00\text{cm}$ for $r_2$ in Equation (V) to find $\left(\Delta V\right)$.

$\begin{array}{c}\left(\Delta V\right)=\left[\frac{\left(8.99\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(-1.6\times {10}^{-19}\text{C}\right)}{\left[\left(2.00\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}\right]-\left[\frac{\left(8.99\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(-1.6\times {10}^{-19}\text{C}\right)}{\left[\left(1.00\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}\right]\\ =7.19\times {10}^{-8}\text{V}\end{array}$

Thus, the potential difference between the points from electron is $\underset{\_}{7.19\times {10}^{-8}\text{V}}$.