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(C) Let the two charges be $Q_1$ and $Q_2$. Given $Q_1 + Q_2 = 36 	imes 10^{-6} \ C$ and $r = 4 \ m$.Using Coulomb's Law: $F = \frac{k Q_1 Q_2}{r^2}$

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$20$

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(C) Let the two charges be $Q_1$ and $Q_2$. Given $Q_1 + Q_2 = 36 	imes 10^{-6} \ C$ and $r = 4 \ m$.Using Coulomb's Law: $F = \frac{k Q_1 Q_2}{r^2}$,where $k = 9 	imes 10^9 \ N \cdot m^2/C^2$.Substituting the values: $0.18 = \frac{9 	imes 10^9 	imes Q_1 	imes Q_2}{4^2}$.$0.18 = \frac{9 	imes 10^9 	imes Q_1 Q_2}{16}$.$Q_1 Q_2 = \frac{0.18 	imes 16}{9 	imes 10^9} = 0.02 	imes 16 	imes 10^{-9} = 320 	imes 10^{-12} \ C^2 = 320 	imes 10^{-12} \ C^2$.We have $Q_1 + Q_2 = 36 	imes 10^{-6}$ and $Q_1 Q_2 = 320 	imes 10^{-12}$.These are roots of the quadratic equation $x^2 - (Q_1+Q_2)x + Q_1 Q_2 = 0$.$x^2 - (36 	imes 10^{-6})x + 320 	imes 10^{-12} = 0$.Solving for $x$: $x = \frac{36 	imes 10^{-6} \pm \sqrt{(36 	imes 10^{-6})^2 - 4(320 	imes 10^{-12})}}{2}$.$x = \frac{36 	imes 10^{-6} \pm \sqrt{1296 	imes 10^{-12} - 1280 	imes 10^{-12}}}{2} = \frac{36 	imes 10^{-6} \pm \sqrt{16 	imes 10^{-12}}}{2}$.$x = \frac{36 	imes 10^{-6} \pm 4 	imes 10^{-6}}{2}$.The two charges are $20 	imes 10^{-6} \ C$ and $16 	imes 10^{-6} \ C$.Thus,the bigger charge is $20 \mu C$.

Two positive point charges are separated by a distance of $4 \ m$ in air. If the sum of the two charges is $36 \mu C$ and the electrostatic force between them is $0.18 \ N$,then the bigger charge is (in $\mu C$)

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