Respuesta :
Answer:
Q must be placed at 0.53 L
Explanation:
Given data:
q_1 = 4.0 μC , q_2 = 3.0μC
Distance between charge is L
third charge q be placed at distance x cm from q1
The force by charge q_1 due to q is
[tex]F1 = \frac{k q q_1}{x^2}[/tex]
[tex]F1 = \frac{k q ( 4.0 μC )}{ x^2}[/tex] ----1
The force by charge q_2 due to q is
[tex]F2 = \frac{k q q_2}{(L-x)^2}[/tex]
[tex]F2 = \frac{kq (3.0 μC)}{(L-x)^2}[/tex] --2
we know that net electric force is equal to zero
F_1 = F_2
[tex]\frac{k q ( 4.0 μC )}{x^2} =\frac{k q ( 3.0 μC )}{(l-x)^2}[/tex]
[tex]\frac{4}{3}*(L-x)^2 = x^2[/tex]
[tex]x = \sqrt{\frac{4}{3}*(L - x)[/tex]
[tex]L-x = \frac{x}{1.15}[/tex]
[tex]L = x + \frac{x}{1.15} = 1.86 x[/tex]
x = 0.53 L
Q must be placed at 0.53 L
