Chapter 16, Problem 24P

To determine

The charge.

Answer to Problem 24P

The charge is $0.020\text{nC}$.

Explanation of Solution

Refer figure 1.

Write an expression for the net horizontal force.

    $\begin{array}{c}\Sigma F_x=T\sin \frac{\theta }{2}-F\\ =0\end{array}$                                                                                           (I)

Here, $\Sigma F_x$ is the net horizontal force, $T$ is the tension, $\theta$ is the angle and $F$  is the force.

Refer figure 1 and rewrite equation (I).

    $\begin{array}{c}\Sigma F_x=T\left(\frac{d/2}{L}\right)-F\\ =0\end{array}$                                                                                              (II)

Here, $d$ is the separation between the charges and $L$ is the length of the thread.

Rewrite equation (II) to find $F$.

    $\begin{array}{c}F=T\left(\frac{d/2}{L}\right)\\ =\frac{Td}{2L}\end{array}$                                                                                                      (III)

Write an expression for the net vertical force.

    $\begin{array}{c}\Sigma F_y=T\cos \frac{\theta }{2}-mg\\ =0\end{array}$                                                                                        (IV)

Here, $\Sigma F_y$ is the net horizontal force, $m$ is the mass and $g$  is the acceleration due to gravity.

Refer figure 1 and rewrite equation (IV).

    $\begin{array}{c}\Sigma F_x=T\left(\frac{\sqrt{L^2-{\left(d/2\right)}^2}}{L}\right)-mg\\ =0\end{array}$                                                                            (V)

Rewrite equation (V) to find $T$.

    $\begin{array}{c}T=\frac{mg}{\frac{\sqrt{L^2-{\left(d/2\right)}^2}}{L}}\\ =\frac{mg}{\sqrt{1-{\left(\frac{d}{2L}\right)}^2}}\end{array}$                                                                                               (VI)

Write an expression for electrostatic force between the charges.

    $F=\frac{k{Q}^2}{d^2}$                                                                                                           (VII)

Here, $k$ is the coulomb’s constant and $Q$ is the charge.

Equate equation (I) and (VII).

    $\begin{array}{c}\frac{k{Q}^2}{d^2}=\frac{Td}{2L}\\ =\left(\frac{mg}{\sqrt{1-{\left(\frac{d}{2L}\right)}^2}}\right)\left(\frac{d}{2L}\right)\\ =\frac{mgd}{\sqrt{4L^2-d^2}}\end{array}$                                                                           (VIII)

Rewrite equation (VIII) to find $Q$.

    $Q=\sqrt{\frac{mg{d}^2}{k\sqrt{4L^2-d^2}}}$                                                                                         (IX)

Conclusion:

Substitute $8.988\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k$, $9.80{\text{m/s}}^2$ for $g$, $0.020\text{m}$ for $d$, $9.0\times {10}^{-8}\text{kg}$ for $m$ and $0.98\text{m}$ for $L$ in equation (IX) to find $Q$.

    $\begin{array}{c}Q=\sqrt{\frac{\left(9.0\times {10}^{-8}\text{kg}\right)\left(9.80{\text{m/s}}^2\right){\left(0.020\text{m}\right)}^2}{\left(8.988\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right)\sqrt{4{\left(0.98\text{m}\right)}^2-{\left(0.020\text{m}\right)}^2}}}\\ =\left(0.020\times {10}^{-9}\text{C}\right)\left(\frac{1\text{nC}}{{10}^{-9}\text{C}}\right)\\ =0.020\text{nC}\end{array}$

Thus, the charge is $0.020\text{nC}$.