91Ó°ÊÓ

First, let's identify the force acting on the positively charged particle as it moves in a circle around the negatively charged particle. Since the particles are oppositely charged, they will experience an electrostatic force. This force can be represented by Coulomb's Law, which states: \(F = k \frac{q_1 q_2}{r^2}\) where \(F\) is the force between the two particles, \(k\) is the electrostatic constant, \(q_1\) and \(q_2\) are the charges of the two particles, and \(r\) is the distance separating them. To determine if work is being done, we need to recall the definition of work. Work is calculated as: \(W = Fd \cos{\theta}\) where \(W\) is the work, \(F\) is the force, \(d\) is the displacement of the object in the direction of force, and \(\theta\) is the angle between the force and displacement vectors. In this scenario, the positively charged particle is moving in a circle, and the electrostatic force is always directed towards the center of the circle. Therefore, the angle between the force and the direction of displacement is always 90 degrees (\(\theta\) = \(90^{\circ}\)). In this case, the cosine of the angle is 0, so the work done equals: \(W = Fd \cos{(90^{\circ})} = Fd \times 0 = 0\) This means that no work is being done on the positively charged particle as it moves in a circle around the negatively charged particle.

In this scenario, an iron nail is being pulled off a magnet. The force acting on the iron nail is a magnetic force. When the iron nail is being pulled off, it experiences a displacement away from the magnet. To determine if work is being done, we need to consider the displacement of the nail and the angle between the force and displacement vectors. In this case, the force vector is in the same direction as the displacement vector, so \(\theta = 0^{\circ}\). The cosine of 0 degrees is 1, so the work done equals: \(W = Fd \cos{(0^{\circ})} = Fd \times 1 = Fd\) Since the force and the displacement are non-zero, work is being done on the iron nail as it is pulled off the magnet.