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Two particles, each carrying positive charge q, are on the y axis, one at y=+a and the other at y=−a. calculate electric field for any point on the x axis

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skyluke89
The electric field produced by a single particle of charge q is given by:
[tex]E(r)=k_e \frac{q}{r^2} [/tex]
where [tex]k_e[/tex] is the Coulomb's constant, and r is the distance of the point we are measuring the field from the particle.

The problem asks to calculate the electric field for any point on the x-axis. Let's start with the charge located at y=+a. Let's take a generic point on the x-axis: this will have coordinates (x,0). Therefore, the distance of this point from the charge will be 
[tex]r= \sqrt{x^2+a^2} [/tex]
And so, if we substitute this into the  formula of the electric field, we have
[tex]E_1 = k_e \frac{q}{x^2+a^2} [/tex]

For the second charge, the electric field is the same. In fact, the  second charge is also +q. It is located at y=-a, therefore the distance from the generic point x will be
[tex]r= \sqrt{x^2+(-a)^2} = \sqrt{x^2+a^2} [/tex]
And so, the field generated by charge 2 is
[tex]E_2=k_e \frac{q}{x^2+a^2} [/tex]

so, the total field is
[tex]E_{tot}(x) = E_1 + E_2=2k_e \frac{q}{x^2+a^2} [/tex]
onyebuchinnaji

The field for any point on the x-axis will determined as [tex]E = \frac{kq}{4a^2+ x^2}[/tex]

The given parameters;

  • charge of each particle, = q
  • the position of one charge = + a
  • position of the second charge = -a

The electric field between these charge is determined by applying Coulomb's law as shown below;

[tex]E = \frac{kq}{r^2}[/tex]

where;

  • r is the distance between the charges

The resultant distance at any point in the x-axis is calculated as follows;

[tex]r^2 = (a- -a)^2 + (x)^2\\\\r^2 = (2a)^2 + (x)^2\\\\r^2 = 4a^2 + x^2[/tex]

The electric field is calculated as follows;

[tex]E = \frac{kq}{r^2} = \frac{kq}{4a^2 + x^2}[/tex]

Thus, the field for any point on the x-axis will determined as [tex]E = \frac{kq}{4a^2+ x^2}[/tex]

Learn more here:https://brainly.com/question/19564329