Two tiny conducting spheres are identical and carry charges of -20.0 μC and +50.0 μC. They are
separated by a distance of 2.50 cm.
(a) What is the magnitude of the force that each sphere experiences, and is the force attractive or repulsive?
(b) The spheres are brought into contact and then separated to a distance of 2.50 cm.
Determine the magnitude of the force that each sphere now experiences, and state whether the force is attractive or repulsive.
To calculate the magnitude of the force that each sphere experiences, we can use Coulomb's Law.
a) Coulomb's Law states that the magnitude of the force between two charged objects is proportional to the product of their charges and inversely proportional to the square of the distance between them. Mathematically, it can be represented as:
F = k * (q1 * q2) / r^2
where F is the force, k is the electrostatic constant (9 x 10^9 N m^2/C^2), q1 and q2 are the charges, and r is the distance between the spheres.
For the sphere with charge -20.0 μC, we'll substitute q1 = -20.0 x 10^-6 C, and for the sphere with charge +50.0 μC, we'll substitute q2 = +50.0 x 10^-6 C. The distance between the spheres is given as 2.50 cm, which we'll convert to meters by dividing by 100 (1 cm = 0.01 m).
Plugging these values into the equation, we have:
F = (9 x 10^9 N m^2/C^2) * (-20.0 x 10^-6 C * +50.0 x 10^-6 C) / (0.025 m)^2
Simplifying this calculation will give us the magnitude of the force that each sphere experiences.
b) After the spheres are brought into contact and separated, they will share their charges to reach an equilibrium state. As a result, the charges on both spheres will become the same (in this case, the average charge of the two spheres). The magnitude of the force between two charged objects depends on the product of their charges, so if the charges are the same, the magnitude of the force will be the same.
Therefore, after they are separated to a distance of 2.50 cm, the magnitude of the force that each sphere experiences will still be the same as calculated in part (a), and it will have the same attractive or repulsive nature as before.
(a) To find the magnitude of the force between the two spheres, we can use Coulomb's law:
Coulomb's Law: F = k * (|q1 * q2|) / r^2
Where:
F is the magnitude of the force
k is the electrostatic constant (k = 8.99 * 10^9 Nm^2/C^2)
q1 and q2 are the charges on the spheres
r is the distance between the centers of the spheres
Using the given information:
q1 = -20.0 μC = -20.0 * 10^-6 C
q2 = +50.0 μC = +50.0 * 10^-6 C
r = 2.50 cm = 2.50 * 10^-2 m
Substituting these values into the formula, we have:
F = (8.99 * 10^9 Nm^2/C^2) * ((|-20.0 * 10^-6 C * +50.0 * 10^-6 C|) / (2.50 * 10^-2 m)^2)
F = (8.99 * 10^9 Nm^2/C^2) * (|-20.0 * 50.0 * 10^-12 C^2|) / (2.50 * 10^-2 m)^2
F = (8.99 * 10^9 Nm^2/C^2) * (1.0 * 10^-9 C^2) / (2.50 * 10^-2 m)^2
F = (8.99 * 1.0) * (1.0 / (2.50*2.50)) * 10^9 N
F ≈ 1.7984 N
Since the two spheres have opposite charges (-20.0 μC and +50.0 μC), the force between them is attractive.
(b) When the spheres are brought into contact, they will share their charges until they have the same charge. In this case, the spheres will have a charge of (-20.0 + 50.0) μC = 30.0 μC.
After the spheres are separated by a distance of 2.50 cm, we can calculate the magnitude of the force using Coulomb's law again:
q1 = -30.0 μC = -30.0 * 10^-6 C
q2 = +30.0 μC = +30.0 * 10^-6 C
Substituting the new charges into the equation, we have:
F = (8.99 * 10^9 Nm^2/C^2) * ((|-30.0 * 10^-6 C * +30.0 * 10^-6 C|) / (2.50 * 10^-2 m)^2)
F = (8.99 * 10^9 Nm^2/C^2) * (|-30.0 * 30.0 * 10^-12 C^2|) / (2.50 * 10^-2 m)^2
F = (8.99 * 10^9 Nm^2/C^2) * (0.9 * 10^-9 C^2) / (2.50 * 10^-2 m)^2
F = (8.99 * 0.9) * (1.0 / (2.50*2.50)) * 10^9 N
F ≈ 0.32304 N
Since the two spheres have the same charge now, the force between them is repulsive.
(a) F=k•q₁•q₂/r²=....
k =9•10⁹ N•m²/C²
attraction
(b) the law of conservation of charge:
q(total) = -20+50 = 30 μC
The charge on each sphere is q=15 μC
F =k•q²/r² = …
repulsion