Physics - Electrostatics Question with Solution | TestHub

Let $k$ be the constant of proportionality in Coulomb's law, and $r$ be the separation between the charges. Since the separation is the same in all cases, we can write the forces as:

$F_A=k\frac{Q_1{q}_3}{r^2}=2$

$F_B=k\frac{Q_2{q}_1}{r^2}=-4$

$F_C=k\frac{Q_3{q}_2}{r^2}=-18$

Also, we have:

$Q_1=q_1+q_2$

$Q_2=q_2+q_3$

$Q_3=q_1+q_3$

Substituting these into the force equations, we get:

$k\frac{(q_1+q_2)q_3}{r^2}=2$ (1)

$k\frac{(q_2+q_3)q_1}{r^2}=-4$ (2)

$k\frac{(q_1+q_3)q_2}{r^2}=-18$ (3)

Let $K=\frac{k}{r^2}$. Then:

$K(q_1+q_2)q_3=2$ (1)

$K(q_2+q_3)q_1=-4$ (2)

$K(q_1+q_3)q_2=-18$ (3)

$K(q_1{q}_3+q_2{q}_3)=2$ (1)

$K(q_1{q}_2+q_1{q}_3)=-4$ (2)

$K(q_1{q}_2+q_2{q}_3)=-18$ (3)

Let $x=q_1{q}_2$, $y=q_2{q}_3$, $z=q_1{q}_3$. Then:

$K(z+y)=2$ (1)

$K(x+z)=-4$ (2)

$K(x+y)=-18$ (3)

Adding the three equations:

$K(2x+2y+2z)=2-4-18=-20$

$K(x+y+z)=-10$ (4)

Subtracting each of the equations (1), (2), and (3) from (4):

$Kx=-10-2=-12$

$Ky=-10-(-4)=-6$

$Kz=-10-(-18)=8$

So, $K{q}_1{q}_2=-12$, $K{q}_2{q}_3=-6$, $K{q}_1{q}_3=8$.

Now, we want to find the ratio $q_1:q_2:q_3$.

We have:

$q_1{q}_2=-\frac{12}{K}$

$q_2{q}_3=-\frac{6}{K}$

$q_1{q}_3=\frac{8}{K}$

Multiplying these three equations:

$(q_1{q}_2)(q_2{q}_3)(q_1{q}_3)=q_1^2{q}_2^2{q}_3^2=\frac{(-12)(-6)(8)}{K^3}=\frac{576}{K^3}$

$(q_1{q}_2{q}_3{)}^2=\frac{576}{K^3}$

$q_1{q}_2{q}_3=\pm \frac{24}{K^{3/2}}$

Now, divide $q_1{q}_2{q}_3$ by each of the pairwise products:

$q_3=\frac{q_1{q}_2{q}_3}{q_1{q}_2}=\frac{\pm 24/K^{3/2}}{-12/K}=\mp \frac{2}{K^{1/2}}$

$q_1=\frac{q_1{q}_2{q}_3}{q_2{q}_3}=\frac{\pm 24/K^{3/2}}{-6/K}=\mp \frac{4}{K^{1/2}}$

$q_2=\frac{q_1{q}_2{q}_3}{q_1{q}_3}=\frac{\pm 24/K^{3/2}}{8/K}=\pm \frac{3}{K^{1/2}}$

Therefore, $q_1:q_2:q_3=\mp \frac{4}{K^{1/2}}:\pm \frac{3}{K^{1/2}}:\mp \frac{2}{K^{1/2}}=-4:3:-2$ or $4:-3:2$

Taking the magnitudes, we have $4:3:2$.

Final Answer: The final answer is $\boxed{4:-3:2}$