Question

in the diagram, q1,q2, and q3 are in a straight line. each of these particles has a charge of -2.35x10^-6 C. particles q1 and q2 are separated by 0.100 m and particles q2 and q3 are separated by 0.100 m. what is the net force on particle q3?

Answer 1

The magnitude of the net force that is acting on particle q₃ is equal to 6.2 Newton.

Given the following data:

Charge = $-2.35 \times 10^{-6}$ C.

Distance = 0.100 m.

Scientific data:

Coulomb's constant = $8.988\times 10^9 \;Nm^2/C^2$

How to calculate the net force.

In this scenario, the magnitude of the net force that is acting on particle q₃ is given by:

F₃ = F₁₃ + F₂₃

Mathematically, the electrostatic force between two (2) charges is given by this formula:

$F = k\frac{q_1q_2}{r^2}$

Where:

• q represent the charge.

• r is the distance between two charges.

• k is Coulomb's constant.

Note: d₁₃ = 2d₂₃ = 2(0.100) = 0.200 meter.

For electrostatic force (F₁₃);

$F_{13} = 8.988\times 10^9 \times \frac{(-2.35 \times 10^{-6} \times [-2.35 \times 10^{-6}])}{0.200^2}\\\\F_{13} = \frac{0.0496}{0.04}$

F₁₃ = 1.24 Newton.

For electrostatic force (F₂₃);

$F_{13} = 8.988\times 10^9 \times \frac{(-2.35 \times 10^{-6} \times [-2.35 \times 10^{-6}])}{0.100^2}\\\\F_{13} = \frac{0.0496}{0.01}$

F₂₃ = 4.96 Newton.

Therefore, the magnitude of the net force that is acting on particle q₃ is given by:

F₃ = 1.24 + 4.96

F₃ = 6.2 Newton.

Read more on charges here: brainly.com/question/14372859