A charged particle A exerts a force of 2.62 N to the right on charged particle B when the particles are
13.7 mm apart. Particle B moves straight away from A to make the distance between them 17.7 mm.
What vector force does particle B then exert on A?
To find the vector force that particle B exerts on particle A, we need to use Coulomb's law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
The equation for Coulomb's law is:
F = k * (|q1| * |q2|) / r^2
Where:
- F is the force between the two particles,
- k is the electrostatic constant (9 x 10^9 N m^2/C^2),
- q1 and q2 are the magnitudes of the charges of the two particles, and
- r is the distance between the particles.
In this case, we know that particle A exerts a force of 2.62 N to the right on particle B when they are 13.7 mm (or 0.0137 m) apart.
Let's assume particle A has a charge qA and particle B has a charge qB. Since particle A exerts a force on particle B, we can write the equation as follows:
2.62 N = k * (|qA| * |qB|) / (0.0137 m)^2
Now, with the given information, we need to find the charge of particle B.
To do that, we rearrange the equation to solve for qB:
|qB| = (2.62 N * (0.0137 m)^2) / (k * |qA|)
Substituting the known values:
|qB| = (2.62 N * (0.0137 m)^2) / (9 x 10^9 N m^2/C^2 * |qA|)
Calculating this expression will give us the magnitude of the charge of particle B. Since the question asks for the vector force, the magnitude of the charge is sufficient.