Chapter 17, Problem 92P

(a)

To determine

What is the potential at point A?

Answer to Problem 92P

The potential at point A is $\underset{\_}{-1.5\text{kV}}$.

Explanation of Solution

Write the equation to find the potential at point A.

    $V=\frac{kQ}{r}$                                                                                                                         (I)

Here, $V$ is the potential, $k$ is a constant, $Q$ is the charge, $r$ is the distance

Conclusion

Substitute $8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k$ , $-50.0\times {10}^{-9}\text{C}$ for $Q$ and $0.30\text{m}$ for $r$ in equation (I) to get $V$.

    $\begin{array}{c}V=\frac{\left(8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(-50.0\times {10}^{-9}\text{C}\right)}{0.30\text{m}}\\ =-1.5\text{kV}\end{array}$

Therefore, the potential at point A is $\underset{\_}{-1.5\text{kV}}$.

(b)

To determine

What is the potential at point B?

Answer to Problem 92P

The potential at point B is $\underset{\_}{-900\text{V}}$.

Explanation of Solution

Write the equation to find the potential at point A.

    $V=\frac{kQ}{r}$                                                                                                                         (II)

Here, $V$ is the potential, $k$ is a constant, $Q$ is the charge, $r$ is the distance

Conclusion:

Substitute $8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k$ , $-50.0\times {10}^{-9}\text{C}$ for $Q$ and $0.50\text{m}$ for $r$ in equation (II) to get $V$.

    $\begin{array}{c}V=\frac{\left(8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(-50.0\times {10}^{-9}\text{C}\right)}{0.50\text{m}}\\ =-900\text{V}\end{array}$

Therefore, the potential at point B is $\underset{\_}{-900\text{V}}$.

(c)

To determine

Through what potential difference does it move if charge is moved from point A to B?

Answer to Problem 92P

The charge moves through a potential of $\underset{\_}{600\text{V}}$ when moved from point A to point B.

Explanation of Solution

Write the equation to find the potential difference between points A and B.

    $\Delta V=kQ\left(\frac{1}{r_B}-\frac{1}{r_A}\right)$                                                                                               (III)

Here, $r_B$ is the distance from charge to point B, and $r_A$ is the distance from charge to point A.

Conclusion:

Substitute $8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k$ , $-50.0\times {10}^{-9}\text{C}$ for $Q$ and $0.50\text{m}$ for $r_B$, and $0.30\text{m}$ for $r_A$ in equation (III) to get $\Delta V$.

    $\begin{array}{c}\Delta V=\left(8.988\times {10}^{-9}{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(-50.0\times {10}^{-9}\text{C}\right)\left(\frac{1}{0.50\text{m}}-\frac{1}{0.30\text{m}}\right)\\ =600\text{V}\end{array}$

Since the potential difference is greater than zero, potential increases.

Therefore, the charge moves through a potential of $\underset{\_}{600\text{V}}$ when moved from point A to point B.

(d)

To determine

What is the change in electric potential energy as it moves from point A to B?

Answer to Problem 92P

As the charge moves from point A to B, its potential energy change is $\underset{\_}{-6.0\times {10}^{-7}\text{J}}$.

Explanation of Solution

Write the equation to find the change in potential energy.

    $\Delta U_E=q\Delta V$                                                                                                         (IV)

Here, $\Delta U_E$ is the change in potential energy, $q$ is the charge, $\Delta V$ is the change in potential

Conclusion:

Substitute $-1.0\times {10}^{-9}\text{C}$ for $q$ and $6.0\times {10}^2\text{V}$ for $\Delta V$ in equation (IV) to get $\Delta U_E$.

      $\begin{array}{c}\Delta U_E=\left(-1.0\times {10}^{-9}\text{C}\right)\left(6.0\times {10}^2\text{V}\right)\\ =-6.0\times {10}^{-7}\text{J}\end{array}$

Therefore, as the charge moves from point A to B, its potential energy change is $\underset{\_}{-6.0\times {10}^{-7}\text{J}}$.

(e)

To determine

How much work is done by the electric field due to charge $Q$ as $q$ moves from A to B?

Answer to Problem 92P

The work done by the electric field due to charge $Q$ as $q$ moves from A to B is $\underset{\_}{6.0\times {10}^{-7}\text{J}}$.

Explanation of Solution

Work done by electric field is equal to the change in electric potential energy due to the motion of charge from point A to point B.

    $W_{field}=-\Delta U_E$                                                                                                        (V)

Here, $W_{field}$ is the work done by the electric field

Conclusion:

Substitute $-6.0\times {10}^{-7}\text{J}$ for $\Delta U_E$ in equation (V) to get $W_{field}$

    $\begin{array}{c}{W}_{field}=-\left(-6.0\times {10}^{-7}\text{J}\right)\\ =6.0\times {10}^{-7}\text{J}\end{array}$

Therefore, The work done by the electric field due to charge $Q$ as $q$ moves from A to B is $\underset{\_}{6.0\times {10}^{-7}\text{J}}$.