Question

Two point charges are 50 m apart. Their combined value is 20 C . What are the magnitudes of the charges if one pepels other with a force of a) 75 N (b) 7.5N (c) 0.75 N

Answer 1

$F=k\frac{q_1q_2}{r^2}$ and $q_1+q_2=20$

So, we have

$F=k\frac{q_1(20-q_1)}{r^2}\to q_1^2-20q_1+\frac{Fr^2}{k}=0$ or

(a)

$q_1^2-20q_1+2.083\cdot 10^{-5}=0 \to q_1=19.99999895834995$ $C$ or

$q_1=0.000001041650053323906$ $C$

$q_2=0.000001041650053323906$ $C$ or $q_2=19.99999895834995$ $C$ accordingly.

(b)

$q_1^2-20q_1+2.083\cdot 10^{-6}=0 \to q_1=19.999999895835$ $C$ or

$q_1=1.0416500018095576\cdot10^{-7}$ $C$

$q_2=1.0416500018095576\cdot10^{-7}$ $C$ or $q_2=19.999999895835$ $C$ accordingly.

(c)

$q_1^2-20q_1+2.083\cdot 10^{-7}=0 \to q_1=19.9999999895835$ $C$ or

$q_1= 1.0416499662824208\cdot 10^{-8}$ $C$

$q_2=1.0416499662824208\cdot 10^{-8}$ $C$ or $q_2=19.9999999895835$ $C$ accordingly.