Question

Two particles are fixed on an x axis. Particle 1 of charge 44.9 μC is cated at х- 24.5 cm; particle 2 of charge Q is located at x 7.53 cm. Particle 3 o charge magnitude 38.9 uc is released from rest on the y axis at y 24.5 cm, what is the value of Q if the initial acceleration of particle 3 is in the positive direction of (a) the x axis and (b) the y axis?

Answer 1

The force that is being exerted on particle 3 by particle 1 is equal to:

$F_{13}=\frac{K*Q_1*Q_3}{r13^2}$

$r_{13} = \sqrt{(0.245m)^2 + (0.245m)^2}=0.3465m^2$

$F_{13}=\frac{9*10^9Nm^2/C^2*44.9*10^-6C*38.9*10^-6C}{0.3465m^2}=45.37N$

As both particles has positive charges, the particles will repel each other and the resultant force will have the direction of the vector $r_{13}$. Therefore, $F_{13}$ will have x and y components equal to:

$F{13x}=F{13}*\frac{r13x}{|r13|}=45.37N*\frac{0.245m}{0.3465m}=32.08 N$

$F{13y}=F{13}*\frac{r13y}{|r13|}=45.37N*\frac{0.245m}{0.3465m}=32.08 N$

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.

$F_{23}=\frac{K*Q_2*Q_3}{r13^2}$

$r_{23} = \sqrt{(-0.0753m)^2 + (0.245m)^2}=0.2563m^2$

$F_{23}=\frac{9*10^9Nm^2/C^2*Q_2*38.9*10^-6C}{0.2563m^2}=1.36*10^6Q_2$

$F{23x}=F{23}*\frac{r23x}{|r23|}=1.36*10^6Q_2*\frac{-0.0753m}{0.2563m}= -401286.15Q_2$

$F{23y}=F{23}*\frac{r23y}{|r23|}=1.36*10^6Q_2*\frac{0.245m}{0.2563m}=1.305*10^6 Q_2$

a) For incise a, F13y + F23y has to be equal to 0:

$F{13y}+F{23y}=0$

$32.08 N+1.305*10^6 Q_2=0$

$Q_2=\frac{-32.08}{1.305*10^6}=-24.6*10^{-6}C =-24.6uC$

b) For incise b, F13x + F23x has to be equal to 0:

$F{13x}+F{23x}=0$

$32.08 N - 401286.15 Q_2=0$

$Q_2=\frac{32.08}{401286.15}=80*10^{-6}C =80uC$