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:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

$F_{23} = \frac{k*q_{2}*q_{3}}{x^2}$ (Electric force of q₂ over q₃)

$F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2}$ (Electric force of q₁ over q₃)

Known data

q₁ = 15 μC = 15*10⁻⁶ C

q₂ = 6 μC = 6*10⁻⁶ C

Problem development

F₂₃ = F₁₃

$\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2}$ (We cancel k and q₃)

$\frac{q_2}{x^2}=\frac{q_1}{(2-x)^2}$

q₂(2-x)² = q₁x²

6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)

6(2-x)² = 15(x)²

6(4-4x+x²) = 15x²

24 - 24x + 6x² = 15x²

9x² + 24x - 24 = 0

The solution of the quadratic equation is:

x₁ = 0.775m

x₂ = -3.44m

x₁ meets the conditions for the forces to cancel in q₃

x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel

The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.

Answer 2

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.