Respuesta :
Answer:
The third charge be placed at a distance of 30 cm from first charge such that the net electric force on it is zero.
Explanation:
Given that,
Charge 1, [tex]q_1=-6\times 10^{-9}\ C[/tex]
Charge 2, [tex]q_2=-3\times 10^{-9}\ C[/tex]
The distance between charges, d = 60 cm
Let the third charge is placed at a distance of x from first charge. So, it will be at a distance of (0.6-x) from charge 2. So,
[tex]k\dfrac{q_1q_2}{x^2}=k\dfrac{q_1q_2}{(60-x)^2}\\\\\dfrac{1}{x^2}=\dfrac{1}{(60-x)^2}\\\\\dfrac{1}{x^2}=\dfrac{1}{3600+x^2-120x}\\\\x=30\ cm[/tex]
So, the third charge be placed at a distance of 30 cm from first charge such that the net electric force on it is zero.
The position of this third charge when the net electric force is zero is 30 cm.
The given parameters:
- First charge, q1 = -6.0 x 10⁻⁹ C
- Second charge, q₂ = -3.0 x 10⁻⁹ C
- Distance between the two charges, r = 60.0 cm
Let the position of the third charge from the first charge = x
The position of this third charge when the net electric force is zero is calculated as follows;
[tex]\frac{kq_1q_2 }{x^2} = \frac{kq_1q_2}{(60-x)^2} \\\\\frac{1}{x^2} = \frac{1}{(60-x)^2} \\\\x^2 = (60-x)^2\\\\x^2 = 3600 - 120x + x^2\\\\120x = 3600\\\\x = \frac{3600}{120} \\\\x = 30 \ cm[/tex]
Thus, the position of this third charge when the net electric force is zero is 30 cm.
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