91影视

Charge $q_{1}isatx=-a$

Charge $q_{2}isatx=+a$

The net force on the third particle of charge -Q is to be zero.

Using the given data of separation, the ratio of the charges can be found easily by substituting these values in the formula of force obtained from Coulomb's law.

Formula:

The magnitude of the electrostatic force between any two particles,

$F_1=\frac{1}{\left(4蟺{蔚}_0\right)}\frac{\left|q_1\right|\left|q_2\right|}{r^2}$ (1)

The net force from particles 1 and 2 on third particle 3 (acting on -Q) is zero. Thus, the relation for x=+0.500a is given using equation (1) as:

$$\begin{gathered} \frac{1}{\left(4蟺{蔚}_0\right)}\frac{\left|q_1\right|Q}{(-a-{\displaystyle \frac{a}{2}}{)}^2}=\frac{1}{\left(4蟺{蔚}_0\right)}\frac{\left|q_2\right|Q}{(a-{\displaystyle \frac{a}{2}}{)}^2} \\ \left|q_1\right|=9.0\left|q_2\right| \\ \frac{\left|q_1\right|}{\left|q_2\right|}=9.0 \end{gathered}$$

Since Q is located between q₁ and q₂, we conclude q₁and q₂ are like-sign.

Hence, the value of the ratio is 9.0.

The net force from particles 1 and 2 on third particle 3 (acting on -Q) is zero. Thus, the relation for x=+1.50a is given using equation (1) as:

$$\begin{gathered} \frac{1}{\left(4蟺{蔚}_0\right)}\frac{\left|q_1\right|Q}{(-a-{\displaystyle \frac{3a}{2}}{)}^2}=\frac{1}{\left(4蟺{蔚}_0\right)}\frac{\left|q_2\right|Q}{(a-{\displaystyle \frac{3a}{2}}{)}^2} \\ \left|q_1\right|=25\left|q_2\right| \\ \frac{\left|q_1\right|}{\left|q_2\right|}=25 \end{gathered}$$

Now, Q is not located between q₁ and q₂ ; one of them must push and the other must pull. Thus, they are unlike-sign, the value of the ratio is -25