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• Notes

Two point charges, $q_{1}=+25 \mathrm{nC}$ and $q_{2}=-75 \mathrm{nC}$ , are sepa-rated by a distance $r=3.0 \mathrm{cm}$ (Fig. 21.12 a) . Find the magnitude
and direction of the electric force (a) that $q_{1}$ exerts on $q_{2}$ and (b) that $q_{2}$ exerts on $q_{1}$ .

$q_{1}=+2 s n c, \quad q_{2}=75 n c, \quad r=3 \mathrm{cm}$

(a) $F_{1 on 2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\left|q, q_{2}\right|}{r^{2}}$

$=9+10^{9} * \frac{\left|\left(25 * 10^{-9}\right)\left(-75 * 10^{-7}\right)\right|}{(0 \cdot 03)^{2}}=0.019 N$

(b) $F_{2 on 1}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\left|q_{1} q_{2}\right|}{1^{2}}=F_{1 on 2}=0.019 n$

A negative charge of $-0.550 \mu \mathrm{C}$ exerts an upward $0.200-\mathrm{N}$ force on an unknown charge 0.300 $\mathrm{m}$ directly below it. (a) What is
the unknown charge (magnitude and sign)? (b) What are the mag-nitude and direction of the force that the unknown charge exerts on
the $-0.550-\mu \mathrm{C}$ charge?

$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{|q_{1} q_2|}{1^2}$

$0.2=9 * 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{c}^{2} * \frac{\left|\left(-0.55*10^{-6}\right)\right|\left|q_{2}\right|}{(0.3)^{2}}$

$\left|q_{2}\right|=+3.64 * 10^{-6} c \quad \rightarrow \quad q_{1}<0, q_{2}=+3.64 * 10^{-6} c$

(b) $F=0.2 \mathrm{N}$ (down word)