Question

Two charges, one of 2.50μC and the other of -3.50μC, are placed on the x-axis, one at the origin and the other at x = 0.600 m
Find the position on the x-axis where the net force on a small charge +q would be zero.
Express your answer in meters.

Answer 1

The electric force between the two bodies which is set at the rest is directly proportional to the product of the charges and inversely proportional to the square of the distance between them

The position on the x-axis where the net force on a small charge +q would be zero is 3.27 meters.

Given information-

The value of first charge is 2.50 μC.

The value of second charge is -3.50 μC.

The position of first charge is at origin.

The position of second charge is at 0.600 m on x-axis.

The position on the x-axis where the net force on a small charge +q would be zero. Thus,

$F_{q1}+F_{q2}=0$

What is force charge?

The electric force between the two bodies which is set at the rest is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Thus,

$k\dfrac{2.5q}{x^2} +k\dfrac{-3.5q}{(x+0.6)^2} =0$

Here $k$ is the column's constant

$\begin{aligned}\\k\dfrac{2.5q}{x^2} -k\dfrac{3.5q}{(x+0.6)^2} &=0\\k\dfrac{2.5q}{x^2} &=k\dfrac{3.5q}{(x+0.6)^2}\\\dfrac{(x+0.6)^2}{x^2} &=\dfrac{3.5}{2.5}\\ \dfrac{(x+0.6)}{x} &=\sqrt{\dfrac{3.5}{2.5}} \\x+0.6&=1.1832x\\x&=3.27\\\end$

Hence the position on the x-axis where the net force on a small charge +q would be zero is 3.27 meters.

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