Chapter 17, Problem 94P

(a)

To determine

What is the change in the electric potential energy.

Answer to Problem 94P

The change in electrical potential energy is $\underset{\_}{0\text{J}}$.

Explanation of Solution

Write the equation to find the electric potential energy at point b.

    $U_b=q\left(\frac{k{q}_L}{r_L}+\frac{k{q}_R}{r_R}\right)$                                                                                                   (I)

Here, $q$ is the charge, $q_L$ is the charge at left side , $q_R$ is the charge at right side, $r_L$ is the distance between the left and central charge, $r_R$ is the distance between charge on right side and central charge.

Both charges are of equal magnitude but sign is opposite.

    $q_L=-q_R$

Substitute $-q_L$ for $q_R$ in equation (I) and simplify to get $U_b$

    $U_b=kq{q}_L\left(\frac{1}{r_L}-\frac{1}{r_R}\right)$                                                                                                         (II)

Similarly find the potential energy at point c.

Rewrite equation (II) to find $U_c$

    $U_C=kq{q}_L\left(\frac{1}{r_L}-\frac{1}{r_R}\right)$                                                                                                          (III)

Take the difference between equation (II) and (III) to find the difference in potential energy.

    $\Delta U=U_b-U_C$                                                                                                               (IV)

Conclusion

Substitute $8.8988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k$ , $-4.2\text{nC}$ for $q$ , $10.0\text{nC}$ for $q_L$ , $4.00\text{cm}$ for $r_L$ and $r_R$ in equation (II) to get $U_b$

    $\begin{array}{c}{U}_b=\left(8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(-4.2\text{nC}\left(\frac{1\text{C}}{{10}^9\text{nC}}\right)\right)\left(\frac{1}{4.00\text{cm}\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)}-\frac{1}{4.00\text{cm}\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)}\right)\\ =0\text{J}\end{array}$

Substitute $8.8988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k$ , $-4.2\text{nC}$ for $q$ , $10.0\text{nC}$ for $q_L$ , $8.00\text{cm}$ for $r_L$ and $r_R$ in equation (II) to get $U_b$

    $\begin{array}{c}{U}_b=\left(8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(-4.2\text{nC}\left(\frac{1\text{C}}{{10}^9\text{nC}}\right)\right)\left(\frac{1}{8.00\text{cm}\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)}-\frac{1}{8.00\text{cm}\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)}\right)\\ =0\text{J}\end{array}$

Substitute $0\text{J}$ for $U_b$ and $U_c$ in equation (IV) to get $\Delta U$

    $\begin{array}{c}\Delta U=0\text{J}-0\text{J}\\ \text{=0}\text{J}\end{array}$

Therefore, the change in electrical potential energy is $\underset{\_}{0\text{J}}$.

(b)

To determine

What is the work required to move charge from point a to point b.

Answer to Problem 94P

The work required to move charge from point a to b is $\underset{\_}{-6.3\text{μJ}}$.

Explanation of Solution

Work done is equal to the change in electrical potential energy.

Write the equation to find the potential energy at point a.

    $U_a=q\left(\frac{k{q}_L}{r{L}_{}}+\frac{k{q}_R}{r_R}\right)$                                                                                                            (V)

Here, $q$ is the charge, $q_L$ is the charge at left side , $q_R$ is the charge at right side, $r_L$ is the distance between the left and central charge, $r_R$ is the distance between charge on right side and central charge.

Both charges are of equal magnitude but sign is opposite.

    $q_L=-q_R$

Substitute $-q_L$ for $q_R$ in equation (V) and simplify to get $U_b$

    $U_b=kq{q}_L\left(\frac{1}{r_L}-\frac{1}{r_R}\right)$                                                                                                        (VI)

Write the equation to find the work done.

    $W=U_a-U_b$                                                                                                                (VII)

Conclusion:

Substitute $8.8988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k$ , $-4.2\text{nC}$ for $q$ , $10.0\text{nC}$ for $q_L$ , $4.00\text{cm}$ for $r_L$ and $12.0\text{cm}$ for $r_R$ in equation (VI) to get $U_a$

    $\begin{array}{c}{U}_a=\left(8.988\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(-4.2\text{nC}\left(\frac{1\text{C}}{{10}^9\text{nC}}\right)\right)\left(\frac{1}{4.00\text{cm}\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)}-\frac{1}{12.00\text{cm}\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)}\right)\\ =-6.3\text{μJ}\end{array}$

Substitute $-6.3\text{μJ}$ for $U_a$ and $0\text{J}$ for $U_b$ in equation (VII) to get $W$

    $\begin{array}{c}W=-6.3\text{μJ}-0\text{J}\\ \text{=}-6.3\text{μJ}\end{array}$

Therefore, The work required to move charge from point a to b is $\underset{\_}{-6.3\text{μJ}}$.