# Two small positively charged spheres have a combined charge of 5.0×10^-5 C. If each sphere is repelled from the other by an electrostatic force of 1.0N when the spheres are 2.0m apart, what is the charge on each sphere?

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23
Mar 9, 2016

q_1 = 3 .84×10^-5 C
q_2 = 1 .16×10^-5 C

#### Explanation:

We are are not given the values of the individual charges; let them be q1 and q2. The condition on the combined charge of the spheres gives us: q_1 + q_2 =5.0×10^-5 C
The next condition concerns the electrostatic force, and so it involves Coulomb’s Law. Both charges are positive because their sum is positive and they repel each other, thus
$| q 1 | = q 1$ and $| q 2 | = q 2$
Now $F = k \frac{{q}_{1} {q}_{2}}{r} ^ 2 = 1 .0 N$ We know k and r, so we can solve for the value of the product of the charges:
${q}_{1} {q}_{2} = \left(1.0 N\right) {r}^{2} / k$
(1.0N)(2.0m)^2 8.99×10^9 N·m^2 C^ 2 =4 .449×10^-10 C^2
Now we have two equations for the two unknowns q1 and q2.
color(red)(q_2 =5.0×10^-5 −q_1)
${q}_{1} \textcolor{red}{{q}_{2}} = 4.449 \times {10}^{-} 10$
q_1color(red)((5.0×10^-5 −q_1)) = 4.449xx10^-10
(5.0×10^-5q_1 −q_1^2) = 4.449xx10^-10
q_1^2 - (5.0×10^-5 C)q1 +4 .449×10^-10=0
q_(1,2) = ((5×10^-5) +- sqrt((5×10^-5)^2 - 4(4 .449×10^-10 )))/2
q_1 = 3 .84×10^-5 C; q_2 = 1 .16×10^-5 C

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