Question

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 of repulsive.

Answer 1

The magnitude of the force that each conducting sphere experiences can be calculated using Coulomb's Law. The force is attractive when the spheres are separated. After the spheres are brought into contact and then separated, the force is still attractive but with a different magnitude.

Step-by-step explanation:

The magnitude of the force that each conducting sphere experiences can be calculated using Coulomb's Law:

F = k * (|q1 * q2|) / r^2

Where F is the force, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between the charges.

Using the given values, we have:

q1 = -20.0 µC, q2 = +50.0 µC, r = 2.50 cm = 0.025 m

Plugging these values into the equation, we get:

F = 9 * 10^9 * (|-20.0 * 10^-6 * 50.0 * 10^-6|) / (0.025^2)

Solving this equation will give you the magnitude of the force that each sphere experiences.

Since q1 and q2 have opposite signs, the force is attractive.

After the spheres are brought into contact and then separated, each sphere is left with half of the original charge. Therefore, if the magnitude of the force is calculated again using the same formula, you will find that the force is attractive and its magnitude will be different.