Physics question

Question

In the diagram, q1 = -4.60 x 10^-6 C, 92 +3.75 x 10^-5 C, and q3 = -5.30 x 10^-6 C. Find the Y-component of the net force on q2.

The diagram is a right triangle. Q1 is connected to Q3 (up-down) with a distance of 0.200 m. Q3 is connected to Q2 (right-left) with a distance of 0.200 m. Q2 is connected to Q1 (diagonally) with a distance of 0.283 m. Q2 has an interior angle of 45 degrees.

Answer 1

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

F = k * q₁ * q₂ / r²,

where k = 1 / (4 * π * ε₀) in SI units.

The horizontal (x-direction) interaction between q₂ and q₃ can be ignored since we are only interested in forces acting in the y-direction. Since the interaction between q₁ and q₂ is not aligned with the y-direction we will need to find the magnitude of the force between q₁ and q₂ using Coulomb's Law and then project out the y-component with some trigonometry.

The magnitude of the electrostatic force is

F = 1 / [ 4 * (3.14) * (8.85 x 10^-12 C²/Nm²) ] * (-4.6 x 10^-6 C) * (3.75 x 10^-5 C) / (0.283 m)²

= -1.94 x 10¹ N

= -19.4 N.

We can project out our final answer, the y-component of the force, using principles of trigonometry as follows:

F_y = F_{q1 -> q2} * sin(45°)

= (-19.4 N) * (0.707)

= -13.7 N.