Question

Constants 1 Periodic Table Part A Two small nonconducting spheres have a total charge of 95.0 μC When placed 1.11 m apart, the force each exerts on the other is 11.0 N and is repulsve. What is the charge on each? Enter your answers in ascending order separated by comma. 1-2.01 10$7.49M10s Correct ▼PartB What if the force were attractive? Enter your answers in ascending order separated by comma. 44.-1.09 .10 4. 1.4-10 Submit Previows Answers Request Ansr X Incorrect: Try Again

Please answer part 2 thoroughly, I do not understand how my answer is wrong.

Answer 1

Part B)

$Q_1+Q_2=95\,\mu C=95*10^{-6}\,C$

Force between the two charges is attractive, so if $Q_1$ is positive $Q_2$ must be negative.

$F=-11.0\,N$ ( negative sign indicates the force is attractive)

$F=\frac{Q_1Q_2}{4\pi\epsilon_0 r^2}$

$\frac{8.99*10^9\,Q_1Q_2}{(1.11)^2}=-11.0$

$Q_1Q_2=-1.05076*10^{-9}$

$Q_1-Q_2=\sqrt{(Q_1+Q_2)^2-4Q_1Q_2}$

$Q_1-Q_2=\sqrt{(95*10^{-6})^2-4(-1.5076*10^{-9})}=0.0001227$

$Q_1-Q_2=0.0001227$ ---(1)

$Q_1+Q_2=0.000095$ ---(2)

adding equations (1) and (2) $2Q_1=0.0002177 \Rightarrow Q_1=0.000109$ ,

using the above value in (1) $0.000109-Q_2=0.0001227\Rightarrow Q_2=-0.000014$

  $Q_1=1.09*10^{-4}\,C$ and $Q_2=-1.4*10^{-5}\,C$

(Verification:

$Q_1+Q_2=0.000109+(-0.000014)=0.000095=95*10^{-6}\,C$)