Respuesta :
Answer:
3.62 m and - 1.4 m
Explanation:
Consider a location towards the positive side of x-axis beyond the location of charge Q₂
x = distance of the location from charge Q₂
d = distance between the two charges = 2 m
For the electric field to be zero at the location
E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location
[tex]\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}[/tex]
[tex]\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}[/tex]
x = 1.62 m
So location is 2 + 1.62 = 3.62 m
Consider a location towards the negative side of x-axis beyond the location of charge Q₁
x = distance of the location from charge Q₁
d = distance between the two charges = 2 m
For the electric field to be zero at the location
E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location
[tex]\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}[/tex]
[tex]\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}[/tex]
x = - 1.4 m
Electric field is zero due to positive point charge Q1 and negative Q2 along the x axis is at the location of 3.62 meters.
What is electric field?
The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.
Given information-
The charge of the point 1 is [tex]2.5\times10^{-5} \rm C[/tex].
The charge of the point 2 is [tex]-5.0\times10^{-6} \rm C[/tex].
The distance between the point 1 and point 2 is 2 meters away from the x axis.
Let the position of the point 1 is at x.
Thus the electric force on point Q1 is,
[tex]F_1=\dfrac{kQ1}{x^2}[/tex]
As the distance between the point Q1 and point Q2 is 2 meters away from the x axis. Thus the position of it should be at (x+2). The electric force on point 2
[tex]F_2=\dfrac{kQ1}{(x+2)^2}[/tex]
As the force of two is equal and opposite thus,
[tex]\dfrac{kQ1}{x^2}=\dfrac{kQ2}{(x+2)^2}\\\dfrac{2.5\times10^{-5}}{x^2}=\dfrac{-5\times10^{-6}}{(x+2)^2}\\x=1.62[/tex]
Thus the position of point 2 is,
[tex]p_2=2+1.62\\p_2=3.62\rm m[/tex]
Thus, the location of the place along the x axis where the electric field due to these two charges is zero is 3.62 meters.
Learn more about electric field here;
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