Question

Two small nonconducting spheres have a total charge of Q=Q1+Q2= 89.0 μC , Q1<Q2. When placed 32.0 cm apart, the force each exerts on the other is 12.0 N and is repulsive.

a.) What is the charge Q1?

b.) What is the charge Q2?

c.) What would Q1 be if the force were attractive?

d.) What would Q2 be if the force were attractive?

Answer 1

$Q_{1}&plus;Q_{2}=89\mu C$

$Q_{2}=(89-Q_{1})\mu C$

$d=32cm=0.32m$

$F=kQ_{1}Q_{2}/d^{2}$

$12N=9*10^{9}*Q_{1}Q_{2}/(0.32m)^{2}$

$12N=9*10^{9}*Q_{1}*10^{-6}C(89-Q_{1})*10^{-6}C/(0.32m)^{2}$

$12=0.088*Q_{1}(89-Q_{1})$

$136.36=89Q_{1}-Q_{1}^{2}$

${\color{Red} Q_{1}^{2}-89Q_{1}&plus;136.36=0}$

Solve the Quadratic equation using a calculator

(a)ANSWER: ${\color{Red} Q_{1}=1.6\mu C}$

$Q_{2}=(89-Q_{1})\mu C$

(b)ANSWER: ${\color{Red} Q_{2}=87.4\mu C}$

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If the force were attractive ,One of the charge should be negative

since $Q_{1}<Q_{2}$ , $Q_{1}$ is negative

$-Q_{1}&plus;Q_{2}=89\mu C$

$Q_{2}=(89&plus;Q_{1})\mu C$

$d=32cm=0.32m$

$F=kQ_{1}Q_{2}/d^{2}$

$12N=9*10^{9}*Q_{1}Q_{2}/(0.32m)^{2}$

$12N=9*10^{9}*Q_{1}*10^{-6}C(89&plus;Q_{1})*10^{-6}C/(0.32m)^{2}$

$12=0.088*Q_{1}(89&plus;Q_{1})$

$136.36=89Q_{1}&plus;Q_{1}^{2}$

${\color{Red} Q_{1}^{2}&plus;89Q_{1}-136.36=0}$

Solve the Quadratic equation using a calculator

(c)ANSWER: ${\color{Red} Q_{1}=-1.5\mu C}$

$Q_{2}=(89&plus;Q_{1})\mu C$

(d)ANSWER: ${\color{Red} Q_{2}=90.5\mu C}$

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