Question

Illustrated in Fig. 24-2, are two identical balls in vaccuum, each of mass $0.10$ g. They carry identical charges and are suspended by two threads of equal length. At equilibrium they position themselves as indicated. Find the charge on either ball. Consider the ball on the left. It is in equilibrium under three forces: (1) the tension $F_{T}$ in the thread; (2) the force of gravity, $$ m g=\left(1.0 \times 10^{-4} \mathrm{~kg}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=9.8 \times 10^{-4} \mathrm{~N} $$ and (3) the Coulomb repulsion $F_{E}$. Writing $\sum F_{x}=0$ and $\sum F_{y}=0$ for the ball on the left, $$ F_{T} \cos 60^{\circ}-F_{E}=0 \quad \text { and } \quad F_{T} \sin 60^{\circ}-m g=0 $$ From the second equation, $$ F_{T}=\frac{m g}{\sin 60^{\circ}}=\frac{9.8 \times 10^{-4} \mathrm{~N}}{0.866}=1.13 \times 10^{-3} \mathrm{~N} $$ Substituting into the first equation gives $$ F_{E}=F_{T} \cos 60^{\circ}=\left(1.13 \times 10^{-3} \mathrm{~N}\right)(0.50)=5.7 \times 10^{-4} \mathrm{~N} $$ But this is the Coulomb force, $k q q^{\prime} / r^{2}$. Therefore, $$ q q^{\prime}=q^{2}=\frac{F_{E} r^{2}}{k}=\frac{\left(5.7 \times 10^{-4} \mathrm{~N}\right)(0.40 \mathrm{~m})^{2}}{9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}} $$ from which $q=0.10 \mu \mathrm{C}$.

          Illustrated in Fig. 24-2, are two identical balls in vaccuum, each of mass $0.10$ g. They carry identical charges and are suspended by two threads of equal length. At equilibrium they position themselves as indicated. Find the charge on either ball.
Consider the ball on the left. It is in equilibrium under three forces: (1) the tension $F_{T}$ in the thread; (2) the force of gravity,
$$
m g=\left(1.0 \times 10^{-4} \mathrm{~kg}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=9.8 \times 10^{-4} \mathrm{~N}
$$
and (3) the Coulomb repulsion $F_{E}$. Writing $\sum F_{x}=0$ and $\sum F_{y}=0$ for the ball on the left,
$$
F_{T} \cos 60^{\circ}-F_{E}=0 \quad \text { and } \quad F_{T} \sin 60^{\circ}-m g=0
$$
From the second equation,
$$
F_{T}=\frac{m g}{\sin 60^{\circ}}=\frac{9.8 \times 10^{-4} \mathrm{~N}}{0.866}=1.13 \times 10^{-3} \mathrm{~N}
$$
Substituting into the first equation gives
$$
F_{E}=F_{T} \cos 60^{\circ}=\left(1.13 \times 10^{-3} \mathrm{~N}\right)(0.50)=5.7 \times 10^{-4} \mathrm{~N}
$$
But this is the Coulomb force, $k q q^{\prime} / r^{2}$. Therefore,
$$
q q^{\prime}=q^{2}=\frac{F_{E} r^{2}}{k}=\frac{\left(5.7 \times 10^{-4} \mathrm{~N}\right)(0.40 \mathrm{~m})^{2}}{9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}}
$$
from which $q=0.10 \mu \mathrm{C}$.

Added by Christopher B.

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