The force that is being exerted on particle 3 by particle 1 is equal to:
[tex]F_{13}=\frac{K*Q_1*Q_3}{r13^2}[/tex]
[tex]r_{13} = \sqrt{(0.245m)^2 + (0.245m)^2}=0.3465m^2[/tex]
[tex]F_{13}=\frac{9*10^9Nm^2/C^2*44.9*10^-6C*38.9*10^-6C}{0.3465m^2}=45.37N[/tex]
As both particles has positive charges, the particles will repel each other and the resultant force will have the direction of the vector [tex]r_{13}[/tex]. Therefore, [tex]F_{13}[/tex] will have x and y components equal to:
[tex]F{13x}=F{13}*\frac{r13x}{|r13|}=45.37N*\frac{0.245m}{0.3465m}=32.08 N[/tex]
[tex]F{13y}=F{13}*\frac{r13y}{|r13|}=45.37N*\frac{0.245m}{0.3465m}=32.08 N[/tex]
In order to calculate Force between particles 2 and 3, we first assume Q2 to be possitive (if it's negative the result will have a negative value, so this doesn't matter). We follow the same line of reasoning we used to calculate F13, just that Q2 will be unknown.
[tex]F_{23}=\frac{K*Q_2*Q_3}{r13^2}[/tex]
[tex]r_{23} = \sqrt{(-0.0753m)^2 + (0.245m)^2}=0.2563m^2[/tex]
[tex]F_{23}=\frac{9*10^9Nm^2/C^2*Q_2*38.9*10^-6C}{0.2563m^2}=1.36*10^6Q_2[/tex]
[tex]F{23x}=F{23}*\frac{r23x}{|r23|}=1.36*10^6Q_2*\frac{-0.0753m}{0.2563m}= -401286.15Q_2[/tex]
[tex]F{23y}=F{23}*\frac{r23y}{|r23|}=1.36*10^6Q_2*\frac{0.245m}{0.2563m}=1.305*10^6 Q_2[/tex]
a) For incise a, F13y + F23y has to be equal to 0:
[tex]F{13y}+F{23y}=0[/tex]
[tex]32.08 N+1.305*10^6 Q_2=0[/tex]
[tex]Q_2=\frac{-32.08}{1.305*10^6}=-24.6*10^{-6}C =-24.6uC[/tex]
b) For incise b, F13x + F23x has to be equal to 0:
[tex]F{13x}+F{23x}=0[/tex]
[tex]32.08 N - 401286.15 Q_2=0[/tex]
[tex]Q_2=\frac{32.08}{401286.15}=80*10^{-6}C =80uC[/tex]
