Question

equipotential surface a has a potential of 5650 v, while equipotential surface b has a potential of 7850 v. a particle has a mass of 3.45 10-2 kg and a charge of 5.90 10-5 c. the particle has a speed of 2.00 m/s on surface a. a force is applied to the particle by an external agent, and the particle moves to surface b, arriving there with a speed of 3.00 m/s. how much work is done by the external agent in moving the particle from a to b?

Answer 1

The work done by the external agent in moving the particle from surface a to surface b is 0.172 J.

To calculate the work done by the external agent in moving the particle from surface a to surface b, we need to find the change in kinetic energy of the particle. We can use the equation:

Work done = ΔKE = KEb - KEa

Where KE is the kinetic energy of the particle. The kinetic energy can be calculated using the equation:

KE = 0.5 * mass * speed2

Substituting the given values, we can calculate the work:

KEa = 0.5 × 3.45e-2 × (2.00)2 = 0.138 J

KEb = 0.5 × 3.45e-2 × (3.00)2 = 0.310 J

Work done = ΔKE = 0.310 J - 0.138 J = 0.172 J