Answer:
The third particle is between point A and B. It is 12 cm from point B (or 8 cm from point A).
Explanation:
We can find the location of the third charge 1 μC such that its net electrostatic force becomes zero by using the electrostatic force formula:
[tex]\boxed{F=\frac{kq_1q_2}{r^2} }[/tex]
where:
Since all particles are positive particles, then they will repel each other and their forces will have opposite direction as shown in the picture. In order the net electrostatic force becomes 0, the third particle ([tex]\bf q_C[/tex]) has to be placed between [tex]\bf q_A[/tex] and [tex]\bf q_B[/tex]. For the third particle, there will be 2 forces:
Given:
Let:
Then:
Since the net electrostatic force equals to 0, then [tex]\bf F_A=F_B[/tex]
[tex]\begin{aligned}\\F_A&=F_B\\\\\frac{kq_Aq_C}{r_{AC}^2} &=\frac{kq_Bq_C}{r_{BC}^2} \\\\\frac{q_Aq_C}{r_{AC}^2} &=\frac{q_Bq_C}{r_{BC}^2}\\\\\frac{4\times1}{(0.2-x)^2} &=\frac{9\times1}{x^2}\\\\9(0.2-x)^2&=4x^2\\9x^2-3.6x+0.36&=4x^2\\5x^2-3.6x+0.36&=0\end{aligned}[/tex]
Now, we can use the abc formula to find the x:
[tex]\begin{aligned}\\x&=\frac{-b\pm\sqrt{b^2-4ac} }{2a} \\\\&=\frac{-(-3.6)\pm\sqrt{(-3.6)^2-4(5)(0.36)} }{2(5)}\\\\&=\frac{3.6\pm2.4}{10} \\\\&=0.6\ or\ 0.12\end{aligned}[/tex]
Since x ≤ 0.2, which means x ≠ 0.6, then x = 0.12 m or 12 cm. Hence the third particle is 12 cm from point B and 8 cm from point A.
