Question

Three charges lie along the x axis. The positive charge q1 = 15 μC is at x= 2.0 m and the positive charge q2 = 6.0μC is at the origin. Where must the negative charge q3 be placed on the x axis so that the resultant electric force on it is zero?

Answer 1

Answer:

The charge q3 must lie in $x=0.775(m)$.

Explanation:

For the net force to be zero in q3, the positive charges must be pulling with the same force but in opposite directions. So

$F_{13}=F_{23}$,

where $F_{13}$ is the force exerted by q1 on q3 and  $F_{23}$ is the force exerted by q2 on q3.

Recalling the Coulomb's law, we know that the magnitud of the electric force between two charges $q_{a}$ and $q_{b}$ is:

$F=k_{e}\frac{q_{a}q_{b}}{r^{2}}$,

where $k_{e}$ is Coulomb's constant and $r$ is the distance between the charges.

In addition to this, it is also important to remember that like charges repel each other and unlike charges attract.

So, we have:

$F_{13}=F_{23}$,

$k_{e}\frac{q_{1}q_{3}}{(x_{1}-x_{3})^{2}}=k_{e}\frac{q_{2}q_{3}}{(x_{2}-x_{3})^{2}}$,

$\frac{q_{1}}{(x_{1}-x_{3})^{2}}=\frac{q_{2}}{(x_{2}-x_{3})^{2}}$,

and remembering that $x_{2}=0$

$\frac{q_{1}}{(x_{1}-x_{3})^{2}}=\frac{q_{2}}{x_{3})^{2}}$,

$\frac{q_{1}}{q_{2}}(x_{3})^{2}=(x_{1})^{2} -2x_{1}x_{3}+(x_{3})^{2}$,

wich leads us to

$(\frac{q_{1}}{q_{2}}-1)(x_{3})^{2}-(x_{1})^{2} +2x_{1}x_{3}=0$,

$(1.5)(x_{3})^{2}-(4)^{2} +4x_{3}=0$

and this is a quadratic equation.

The solutions to this equation are:

• $x_{1}=0.775m$
• $x_{2}=-3.44m$

This two are solutions of the equation, even so, only one is a correct solution to the problem. The correct answer is

$x_{1}=0.775m$.

This is because q3 must lie between q1 and q2 so the attractive forces cancel each other.

Answer 2

The negative charge Q3 must be placed at 0.775m  between Q1 and Q2 on the x axis. So that the resultant electric force on it is zero.

The Coulomb's law:

The electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the separation distance between the two objects.

$\rm \bold {F = k_e \frac{Qa. Qb}{r^2} }$

As we know the net charge will be zero when positive charges will pull with same fore but in opposite direction.

$\rm \bold { { k_e \frac{Q1. Q3}{(x1-x3)^2} = k_e \frac{Q2. Q3}{(x2-x3)^2 } } } \\\\\rm \bold { { \frac{Q1 }{(x1-x3)^2} = \frac{Q2}{(x2-x3)^2 } } }$

Since x2 = 0

$\bold { { \frac{Q1 }{(x1-x3)^2} = \frac{Q2}{(-x3)^2 } } }$

$\rm \bold { \frac{Q1}{Q2} ( x3)^2 = x1^2 - 2x1 .x3 +x3^2}$

This leads to,

$\rm \bold { (1.5) ( x3)^2 -4^2 -4x3 = 0}$

Solve the quadratic equation, we get 2 values. we ignore the negative one.

x1 =  0.775 m.

Hence we can conclude that the negative charge Q3 must be placed between Q1 and Q2 on the x axis. So that the resultant electric force on it is zero.

To know more about force on charge, refer to the link:

https://brainly.com/question/11516376