Question

Figure shows an arrangement of four charged particles, with angle $\theta ={30.0}^o$ and distance $d=2.00cm$. Particle $2$ has charge $q_2=+8.00\times {10}^{-19}C$; particles $3$ and $4$ have charges $q_3=q_4=-1.60\times {10}^{-19}C$.
What is distance $D$ between the origin and particle $2$ if the net electrostatic force, on particle $1$ due to the other particles is zero?

Answer 1

We note that $\cos \left({30}^o\right)=\frac{1}{2}\sqrt{3}$, so that the dashed line distance in the figure is $r=2d/\sqrt{3}$. The net force on $q_1$ due to the two charge $q_3$ and $q_4$ (with $|q_3|=|q_4|=1.60\times {10}^{-19}C$) on the $y$ axis has magnitude
$2\frac{|q_1{q}_3|}{4\pi {\epsilon }_0{r}^2}\cos \left({30}^o\right)=\frac{3\sqrt{3}|q_1{q}_3|}{16\pi {\epsilon }_0{d}^2}$.
This must be set equal to the magnitude of the force exerted on $q_1$ by $q_2=8.00\times {10}^{-19}C=5.00|q_3|$ in order that its net force be zero:
$\frac{3\sqrt{3}|q_1{q}_3|}{16\pi {\epsilon }_0{d}^2}=\frac{|q_1{q}_2|}{4\pi {\epsilon }_0(D+d{)}^2}\Rightarrow D=d(2\sqrt{\frac{5}{3\sqrt{3}}}-1)=0.9245d$.
Given $d=2.00cm$, this then leads to $D=1.92cm$.