RaySix

Post

Electric Force Between Charged Particles

Created by @nathanedwards

at November 1st 2023, 1:39:09 am.

AP_Physics_2-Questions

#physics

#coulombs-law

#electrostatics

Question:

Two charged particles, labeled as Particle A and Particle B, are placed in a vacuum. Particle A has a charge of +4.0 μC and Particle B has a charge of -6.0 μC. The two particles are initially separated by a distance of 0.5 meters.

a) Calculate the magnitude and direction of the electric force experienced by Particle A due to the presence of Particle B.

b) If Particle B is moved to a new location, now 2.0 meters away from Particle A, calculate the new magnitude and direction of the electric force experienced by Particle A.

Answer:

a) The magnitude of the electric force between two charged particles can be calculated using Coulomb's law:

F = \frac{{k \cdot |\mathit{q}_1 \cdot \mathit{q}_2|}}{{r^2}}

where:

F is the magnitude of the electric force,
k is the electrostatic constant ( $k = 8.99 \times 10^9 \, \text{{N·m}}^2/\text{{C}}^2$ ),
$|\mathit{q}_1 \cdot \mathit{q}_2|$ is the product of the magnitudes of the charges of the two particles,
r is the distance between the charges.

For Particle A and Particle B, the magnitude of the electric force experienced by Particle A, $F_\text{{AB}}$ , can be calculated as:

F_\text{{AB}} = \frac{{k \cdot |\mathit{q}_\text{{A}} \cdot \mathit{q}_\text{{B}}|}}{{r^2}}

Substituting the given values:

F_\text{{AB}} = \frac{{(8.99 \times 10^9 \, \text{{N·m}}^2/\text{{C}}^2) \cdot (4.0 \times 10^{-6} \, \text{{C}}) \cdot (6.0 \times 10^{-6} \, \text{{C}})}}{{(0.5 \, \text{{m}})^2}}

Calculating $F_\text{{AB}}$ gives:

F_\text{{AB}} ≈ 0.431 \, \text{{N}}

Since Particle B has a negative charge and the force is calculated for Particle A due to Particle B, the direction of the electric force is attractive, towards Particle B.

Therefore, the magnitude of the electric force experienced by Particle A due to the presence of Particle B is approximately 0.431 N, acting towards Particle B.

b) When Particle B is moved to a new location, 2.0 meters away from Particle A, the new magnitude of the electric force experienced by Particle A, $F'_\text{{AB}}$ , can be calculated using the same formula as before:

F'_\text{{AB}} = \frac{{k \cdot |\mathit{q}_\text{{A}} \cdot \mathit{q}_\text{{B}}|}}{{r'^2}}

where $r'$ is the new distance between the particles.

Substituting the given values:

F'_\text{{AB}} = \frac{{(8.99 \times 10^9 \, \text{{N·m}}^2/\text{{C}}^2) \cdot (4.0 \times 10^{-6} \, \text{{C}}) \cdot (6.0 \times 10^{-6} \, \text{{C}})}}{{(2.0 \, \text{{m}})^2}}

Calculating $F'_\text{{AB}}$ gives:

F'_\text{{AB}} ≈ 0.054 \, \text{{N}}

Similar to the previous case, the direction of the electric force remains attractive, towards Particle B.

Therefore, the new magnitude of the electric force experienced by Particle A when Particle B is 2.0 meters away is approximately 0.054 N, acting towards Particle B.