Chapter 20, Problem 19P

Two particles, with charges of 20.0 nC and –20.0 nC, are placed at the points with coordinates (0, 4.00 cm) and (0, –4.00 cm) as shown in Figure P20.19. A particle with charge 10.0 nC is located at the origin. (a) Find the electric potential energy of the configuration of the three fixed charges. (b) A fourth particle, with a mass of 2.00 × 10^–13 kg and a charge of 40.0 nC, is released from rest at the point (3.00 cm, 0). Find its speed after it has moved freely to a very large distance away.

To determine

The electric potential energy.

Answer to Problem 19P

The electric potential energy is $\underset{\_}{-4.50\times {10}^{-5}\text{J}}$.

Explanation of Solution

Write the equation for the net potential energy.

  $U=U_{12}+U_{23}+U_{13}$        (I)

Here, $U$ is the net potential energy, $U_{12}$, $U_{23}$, $U_{13}$ are the potential energy to bring the charges.

Write the expression for potential energy to bring 20nC charge.

$U_{12}=q_2\left(V_1\right)$        (II)

Here, $q_2$ is the charge, $V_1$ is the potential.

Write the expression for the potential.

$V_1=\frac{k{q}_1}{r}$        (III)

Here, $k$ is the Coulomb’s constant, $q_1$ is the charge and $r$ is the.

Write the expression for potential energy to bring -20nC charge.

$U_{23}+U_{13}=q_3{V}_2+q_3{V}_1$        (IV)

Here, $q_2,q_3$ is the charge, $V_1$ is the potential.

Write the expression for the potential.

$V_2=\frac{k{q}_2}{r}$        (V)

Here, $k$ is the Coulomb’s constant, $q_2$ is the charge and $r$ is the.

Rewrite the expression from (I) by using (II), (III) (IV), (V).

$U=\left[q_2\left(\frac{k{q}_1}{r}\right)\right]+\left[q_3\left\{\left(\frac{k{q}_2}{r}\right)+\left(\frac{k{q}_1}{r}\right)\right\}\right]$        (VI)

Conclusion:

Substitute, $20.0\text{nC}$ for $q_1$, $10.0n\text{C}$ for $q_2$, $-20.0n\text{C}$ for $q_3$, $8.99\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $\text{4}\text{.00cm}$ for $r$ in Equation (VI) to find $U$.

  $\begin{array}{c}U=\left[\left(10.0\text{nC}\right)\left(\frac{1\times {10}^{-9}\text{C}}{1\text{nC}}\right)\right]\left(\frac{\left(8.99\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left[\left(20.0\text{nC}\right)\left(\frac{1\times {10}^{-9}\text{C}}{1\text{nC}}\right)\right]}{\left[\left(\text{4}\text{.00cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}\right)\\ +\left[\left(-20.0\text{nC}\right)\left(\frac{1\times {10}^{-9}\text{C}}{1\text{nC}}\right)\right]\left\{\begin{array}{l}\left(\frac{\left(8.99\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left[\left(10.0\text{nC}\right)\left(\frac{1\times {10}^{-9}\text{C}}{1\text{nC}}\right)\right]}{\left[\left(\text{4}\text{.00cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}\right)\\ +\left(\frac{\left(8.99\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left[\left(20.0\text{nC}\right)\left(\frac{1\times {10}^{-9}\text{C}}{1\text{nC}}\right)\right]}{\left[\left(\text{4}\text{.00cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}\right)\end{array}\right\}\\ =\left(4.50\times {10}^{-5}\text{J}\right)-\left(4.50\times {10}^{-5}\text{J}\right)-\left(4.50\times {10}^{-5}\text{J}\right)\\ =-\left(4.50\times {10}^{-5}\text{J}\right)\end{array}$

Thus, the electric potential energy is $\underset{\_}{-4.50\times {10}^{-5}\text{J}}$.

To determine

The speed of the third object.

Answer to Problem 19P

The speed of the third object is $3.46\times {10}^4\text{m/s}$.

Explanation of Solution

Write the equation from conservation of energy.

  $qV=\frac{1}{2}m{v}^2$        (VII)

Here, $q$ is the charge, $m$ is the mass, $v$ is the speed and $V$ is the net potential.

Write the expression for the net potential.

$V=V_1+V_2+V_3$        (VIII)

Write the expression for potential.

$V_3=\frac{k{q}_3}{r}$        (IX)

Here, $q_3$ is the charge, $V_1$ is the potential.

Rewrite the expression from (VII) by using (III), (V), (VIII) and (IX).

$q\left[\frac{k{q}_1}{r}+\frac{k{q}_2}{r}+\frac{k{q}_3}{r}\right]=\frac{1}{2}m{v}^2$        (X)

Conclusion:

Substitute, $20.0\text{nC}$ for $q_1$, $10.0n\text{C}$ for $q_2$, $-20.0n\text{C}$ for $q_3$, $40\text{nC}$ for $q$, $2.00\times {10}^{-13}\text{kg}$ for $m$ and $\text{4}\text{.00cm}$ for $r$ in Equation (VI) to find $U$.

  $\begin{array}{c}\frac{\left(8.99\times {10}^9{\text{N-m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)}{\left[\left(\text{4}\text{.00cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right]}\\ \left\{\left[\left(20.0\text{nC}\right)\left(\frac{1\times {10}^{-9}\text{C}}{1\text{nC}}\right)\right]+\left[\left(10.0\text{nC}\right)\left(\frac{1\times {10}^{-9}\text{C}}{1\text{nC}}\right)\right]+\left[\left(-20.0\text{nC}\right)\left(\frac{1\times {10}^{-9}\text{C}}{1\text{nC}}\right)\right]\right\}\\ =\frac{1}{2}\left(2.00\times {10}^{-13}\text{kg}\right)v^2\\ v=\sqrt{\frac{2\left(1.20\times {10}^{-4}\text{J}\right)}{2\times {10}^{-13}\text{kg}}}\\ =3.46\times {10}^4\text{m/s}\end{array}$

Thus, the speed of the third object is $3.46\times {10}^4\text{m/s}$.