Question

Three point charges are arranged on a line. Charge q3=+5.00x10^-9C and is at the origin. Charge q2=-3.00x10^-9C and is at x=+4.00cm. Charge q1 is at x=+2.00cm. What is the magnitude and sign (+ or -) for q1 if the net force on q3 is zero?

Answer 1

Concepts and reason

The concept of Coulomb’s law is required to solve the problem.

First, use Coulomb’s law and determine the magnitude of the forces on charge ${q_3}$ due to the charge ${q_1}$ and charge ${q_2}$ . Then equate the two forces and determine the magnitude of the charge ${q_3}$ . Finally, determine the sign of the charge ${q_3}$ by using the sign of charge ${q_1}$ and charge ${q_2}$ .

Fundamentals

According to Coulomb’s law, the net force acting on a charge $q$ due to another charge Q is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them. The expression for the magnitude of the electric force acting on the charge q is given as,

$F = \frac{{kqQ}}{{{r^2}}}$

Here, k is the Coulomb’s constant and r is the distance between the two charges.

The magnitude of the force acting on charge ${q_3}$ due to the charge ${q_1}$ is given as,

${F_{13}} = \frac{{k{q_3}{q_1}}}{{{r_{13}}^2}}$

Here, ${r_{13}}$ is the distance between the charges ${q_1}$ and ${q_3}$ .

The magnitude of the force acting on charge ${q_3}$ due to the charge ${q_2}$ is given as,

${F_{23}} = \frac{{k{q_3}{q_2}}}{{{r_{23}}^2}}$

Here, ${r_{23}}$ is the distance between the charges ${q_2}$ and ${q_3}$ .

The net force acting on the charge ${q_3}$ is zero if the magnitude of the force acting on the charge ${q_3}$ due to the charge ${q_1}$ is equal to the magnitude of the force due to the charge ${q_2}$ . That is,

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

Substitute $\frac{{k{q_3}{q_1}}}{{{r_{13}}^2}}$ for ${F_{13}}$ and $\frac{{k{q_3}{q_2}}}{{{r_{23}}^2}}$ for ${F_{23}}$ in the equation ${F_{23}} = {F_{13}}$ and solve for ${q_1}$ .

$\begin{array}{c}\\\frac{{k{q_3}{q_2}}}{{{r_{23}}^2}} = \frac{{k{q_3}{q_1}}}{{{r_{13}}^2}}\\\\\frac{{{q_2}}}{{{r_{23}}^2}} = \frac{{{q_1}}}{{{r_{13}}^2}}\\\\{q_1} = \frac{{{q_2}{r_{13}}^2}}{{{r_{23}}^2}}\\\end{array}$

Substitute $3.00 \times {10^{ - 9}}{\rm{ C}}$ for ${q_2}$ , 2.00 cm for ${r_{13}}$ , and 4.00 cm for ${r_{23}}$ in the equation ${q_1} = \frac{{{q_2}{r_{13}}^2}}{{{r_{23}}^2}}$ .

$\begin{array}{c}\\{q_1} = \frac{{\left( {3.00 \times {{10}^{ - 9}}{\rm{ C}}} \right){{\left( {2.00{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right)}^2}}}{{{{\left( {4.00{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right)}^2}}}\\\\ = 7.5 \times {10^{ - 10}}{\rm{ C}}\\\end{array}$

The following figure shows the direction of the forces acting on the charge ${q_3}$ .

The charge ${q_3}$ is positive and charge ${q_2}$ is negative. Therefore, the force between the charges ${q_3}$ and ${q_2}$ is attractive and the direction of the force on charge ${q_3}$ due to the charge ${q_2}$ is towards right.

The net force acting on the charge ${q_3}$ is zero if the force acting on the charge ${q_3}$ due to the charge ${q_1}$ is equal and opposite to the force due to the charge ${q_2}$ . Therefore, the direction of the force on charge ${q_3}$ due to the charge ${q_1}$ is towards left. This implies that the force between the charges ${q_1}$ and ${q_3}$ is repulsive.

The force between the charges ${q_1}$ and ${q_3}$ is repulsive if both the charges ${q_3}$ and ${q_1}$ are of same sign. Hence, the sign of the charge ${q_1}$ is positive.

Ans:

The magnitude of the charge ${q_1}$ is $7.5 \times {10^{ - 10}}{\rm{ C}}$ and it is of positive sign.