Question

A small particle has charge -5.00 mC and mass 2.00 * 10-4 kg. It moves from point A, where the electric po- tential is V = +200 V, to point B, where the electric potential A is VB = +800 V. The electric force is the only force acting on the particle. The particle has speed 5.00 m>s at point A. What is its speed at point B? Is it moving faster or slower at B than at A? Explain.

Answer 1

The particle's speed at point B is undefined because it does not have enough energy to reach that point. Therefore, it is moving slower at B than at A.

Step-by-step explanation:

In order to find the final speed of the particle at point B, we can use the conservation of energy principle. The change in potential energy of the particle is equal to the change in its kinetic energy. The initial potential energy at point A is given by:

UA = qVA = (-5.00 mC)(+200 V) = -1000 mJ

The final potential energy at point B is given by:

UB = qVB = (-5.00 mC)(+800 V) = -4000 mJ

Since there is no change in gravitational potential energy, the change in the particle's kinetic energy is equal to the difference in potential energy:

ΔK = ΔU = UB - UA = -4000 mJ - (-1000 mJ) = -3000 mJ

Since the final speed at point B is determined by the kinetic energy, we can solve for it by using the equation:

K = 0.5mv^2

Where K is the kinetic energy, m is the mass of the particle, and v is its velocity. Rearranging the equation, we have:

v = √(2K/m)

Substituting the values we have, we get:

v = √(2(-3000 mJ) / (2.00 × 10^-4 kg)) = √(-30 J/kg) = undefined

Since the final speed is undefined, it means that the particle does not have enough energy to reach point B. Therefore, it is moving slower at B than at A.