Answer:
The equilibrium position for the third charge is 69.28 cm
Explanation:
Given;
q₁ = -5.00 x 10⁻⁹ C
q₂ = -2.00 x 10⁻⁹ C
q₃ = 15.00 x 10⁻⁹ C
distance between q₁ and q₂ = 40.0 cm = 0.4 m
(-q₁)--------------------------------------(-q₂)---------------------------------(+q₃)
At equilibrium the repulsive force between q₁ and q₂ must be equal to attractive force between q₂ and q₃
According to Coulomb's law, repulsive or attractive force between charges is calculated as;
[tex]F = \frac{Kq_1q_2}{r_1^2} = \frac{Kq_2q_3}{r_2^2}[/tex]
where;
F is repulsive or attractive force between charges
K is Coulomb's constant = 8.99 x 10⁹ Nm²/c²
r₁ is the distance between q₁ and q₂
q₁, q₂ and q₃ are the charge
distance between q₂ and q₃, r₂ is calculated as;
[tex]\frac{Kq_1q_2}{r_1^2} = \frac{Kq_2q_3}{r_2^2}\\\\\frac{q_1q_2}{r_1^2} = \frac{q_2q_3}{r_2^2}\\\\r_2^2= \frac{r_1^2q_2q_3}{q_1q_2}\\\\r_2^2= \frac{r_1^2q_3}{q_1} = \frac{0.4^2*15*10^{-9}}{5*10^{-9}} = 0.48\\\\r_2 = \sqrt{0.48} = 0.6928 \ m[/tex]
Therefore, the equilibrium position for the third charge is 69.28 cm
The equilibrium position for a third charge will be "69.28 cm". To understand the calculation, check below.
According to the question,
Charges, q₁ = -5.00 × 10⁻⁹ C
q₂ = -2.00 × 10⁻⁹ C
q₃ = 15.00 × 10⁻⁹ C
Distance b/w q₁ and q₂, r₁ = 40.0 cm or,
= 0.4 m
Coulomb's constant, K = 8.99 × 10⁹ Nm²/c²
By using Coulomb's Law,
→ Attractive force, F = [tex]\frac{K q_1 q_2}{r_1^2}[/tex] or,
= [tex]\frac{K q_2 q_3}{r_2^2}[/tex]
Now,
→ [tex]\frac{q_1 q_2}{r_1^2}[/tex] = [tex]\frac{q_2 q_3}{r_2^2}[/tex]
r₂² = [tex]\frac{r_1^2 q_2 q_3}{q_1 q_2}[/tex]
= [tex]\frac{r_1^2 q_3}{q_1}[/tex]
By substituting the values,
= [tex]\frac{(0.4)^2\times 15\times 10^{-9}}{5\times 10^{-9}}[/tex]
= 0.48
r = √0.48
= 0.6928 m or,
= 69.28 cm
Thus the above approach is correct.
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