Given Data:
Masses
$$
m_{1}=m_{2}=m=10 \mathrm{~g}=0.01 \mathrm{~kg}
$$
Charges
$$
\begin{aligned}
q_{1} & =q_{2}=q=20 \mu C \\
& =20 \times 10^{-6} C
\end{aligned}
$$
Distance
$$
r=10 \mathrm{~cm}=0.1 \mathrm{~m}
$$
To Determine:
$$
\frac{F_{e}}{F_{g}}=\text { ? }
$$ Calculations:
$$
\begin{aligned}
& \frac{F_{e}}{F_{g}}=\frac{\left(k \frac{q_{1} q_{2}}{r^{2}}\right)}{\left(G \frac{m_{1} m_{2}}{r^{2}}\right)} \\
&=\frac{k q_{1} q_{2}}{G m_{1} m_{2}} \\
&=\frac{k q^{2}}{G m^{2}} \\
&=\frac{9 \times 10^{9} \times\left(20 \times 10^{-6}\right)^{2}}{6.67 \times 10^{-11} \times(0.01)^{2}} \\
&=5.4 \times 10^{14}
\end{aligned}
$$
12.2 Calculate vectorially the net electrostatic force on $q$ as shown in the figure.
\section{Given Data:}
$$
\begin{gathered}
q_{1}=1 \mu C=1 \times 10^{-6} C \\
q_{2}=-1 \mu C=-1 \times 10^{-6} C \\
q=4 \mu C=4 \times 10^{-6} C
\end{gathered}
$$
To Determine:
Total Force on $q$
$$
\mathrm{F}=\text { ? }
$$
\section{Calculations:}
From Fig.
$$
\begin{gathered}
\tan \theta=\frac{0.8}{0.6} \\
\Rightarrow \theta=\tan ^{-1}\left(\frac{0.8}{0.6}\right)=53^{\circ}
\end{gathered}
$$
Force Exerted by Charge $q_{1}$ on $q$ :
$$
\begin{gathered}
F_{1}=k \frac{q q_{1}}{r^{2}} \\
=9 \times 10^{9} \times \frac{4 \times 10^{-6} \times 1 \times 10^{-6}}{(1)^{2}} \\
=36 \times 10^{-3} \mathrm{~N}
\end{gathered}
$$
Force Exerted by Charge $q_{2}$ on $q$ :
$$
\begin{gathered}
F_{2}=k \frac{q q_{2}}{r^{2}} \\
=9 \times 10^{9} \times \frac{4 \times 10^{-6} \times 1 \times 10^{-6}}{(1)^{2}} \\
=36 \times 10^{-3} N
\end{gathered}
$$
Now
$$
\begin{gathered}
F_{x}=F_{1 x}+F_{2 x} \\
=F_{1} \cos \theta+F_{2} \cos \theta \\
=36 \times 10^{-3} \times \cos 53^{\circ}+36 \times 10^{-3} \times \cos \left(2 \pi-53^{\circ}\right) \\
=0.043 \mathrm{~N}
\end{gathered}
$$
And
$$
\begin{gathered}
F_{y}=F_{1 y}+F_{2 y} \\
=F_{1} \sin \theta+F_{2} \sin \theta \\
=36 \times 10^{-3} \times \sin 53^{\circ}+36 \times 10^{-3} \times \sin \left(2 \pi-53^{\circ}\right) \\
=0 \mathrm{~N}
\end{gathered}
$$
Magnitude of Resultant Force
$$
\begin{gathered}
F=\sqrt{F_{x}^{2}+F_{y}^{2}} \\
=\sqrt{(0.043)^{2}+(0)^{2}} \\
=0.043 \mathrm{~N}
\end{gathered}
$$
Direction of Resultant Force
$$
\begin{gathered}
\tan \theta=\frac{F_{y}}{F_{x}} \\
\Rightarrow \theta=\tan ^{-1}\left(\frac{0}{0.043}\right)=0^{\circ}
\end{gathered}
$$
(Resultant is along $\mathrm{x}$-axis)
$$
\vec{F}=0.043 \hat{\imath}
$$
12.3 A point charge $q=-8.0 \times 10^{-8} C$ is placed at the origin. Calculate electric field at apoint $2.0 \mathrm{~m}$ from the origin on the z-axis.
\section{Given Data:}
Charge
$$
q=-8.0 \times 10^{-8} C
$$
Distance
$$
r=2 m
$$
Direction: z-axis
$$
\hat{r}=\hat{k}
$$
\section{To Determine:}
Electric Field
$$
E=?
$$
\section{Calculations:}
$\overrightarrow{\mathrm{E}}=k \frac{q}{r^{2}} \hat{k}$
$=9 \times 10^{9} \times \frac{\left(-8.0 \times 10^{-8}\right)}{(2)^{2}} \hat{k}$
$=\left(-1.8 \times 10^{2} \hat{k}\right) N C^{-1}$
12.4 Determine the electric field at the position $\overrightarrow{\mathbf{r}}=(4 \hat{\imath}+3 \hat{\jmath}) m$ caused by a point chargeq $=5.0 \times 10^{-6} C$ placed at origin.
\section{Given Data:}
Position Vector
$$
\overrightarrow{\mathrm{r}}=(4 \hat{\imath}+3 \hat{\jmath}) m
$$
Charge
$$
q=5.0 \times 10^{-6} \mathrm{C}
$$
To Determine:
Electric Field
$$
\overrightarrow{\mathrm{E}}=?
$$
\section{Calculations:}
$\overrightarrow{\mathrm{E}}=k \frac{q}{r^{2}} \hat{r}---(1)$
Now $r=|r|=\sqrt{4^{2}+3^{2}}=5$, and $\hat{r}=\frac{\vec{r}}{|r|}=\frac{4 \hat{\imath}+3 \hat{\jmath}}{5}$
Equation (1) becomes:
$$
\begin{gathered}
\overrightarrow{\mathrm{E}}=k \frac{q}{r^{2}} \hat{r} \\
=9 \times 10^{9} \times \frac{5.0 \times 10^{-6}}{(5)^{2}} \times \frac{4 \hat{\imath}+3 \hat{\jmath}}{5} \\
=360 \times(4 \hat{\imath}+3 \hat{\jmath}) \\
=1440 \hat{\imath}+1080 \hat{\jmath}
\end{gathered}
$$
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