Question

Two small, positively charged spheres have a combined charge of 12.0 × 10-5 C. If each sphere is repelled from the other by an electrostatic force of 0.800 N when the spheres are 1.70 m apart, what is the charge on the sphere with the smaller charge?

Answer 1

Answer:

$q=2.2x10^{-6} C$

Explanation:

Combined charges means that $q1+q2=12.0x10^{-5}C$

The electric force can be found by Coulomb´s law $F=K\frac{q1.q2}{r^{2} }$, where K is Coulomb´s constant, q1 and q2 are charges and r is the distance between both spheres.

$q1.q2=\frac{Fr^{2} }{K} =\frac{0.800N.(1.7m)^{2} }{9x10^{9}\frac{Nm^{2} }{C^{2} }}=2.57x10^{-10}C^{2}$

$q1=\frac{2.57x10^{-10} }{q2}$

replacing $\frac{2.57x10^{-10}}{q2} +q2=12x10^{-5}$

$-q_{2} ^{2} +12x10^{-5}q_{2}-2.57x10^{-10}=0$

Using cuadratic equation to solve q2:

$q2=2.2x10^{-6}C \\ and\\ q2=1.178x10^{-4}C$

Knowing that $q1+q2=12.0x10^{-5}C$

$q1=1.178x10^{-4}C \\ and\\ \\ q1=2.2x10^{-6}C$

Either q1 or q2 the smallest charge is $q=2.2x10^{-6} C$