A small particle has charge −5.00μC and mass 2.00×10−4kg. It moves from point A, where the electric potential is VA=+200V, to point B, where the electric potential is VB=+800V. 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.

(a). The speed at point B is 7.41 m/s.

(b). It is moving faster at B.

Explanation:

Given that,

Charge = -5.00μC

Mass of particle

Potential at point A= 200 V

Potential at point B= 800 V

Speed of particle at point A = 5.00 m/s

We need to calculate the speed at point B

Using law of conservation of energy

Put the value into the formula

Hence, (a). The speed at point B is 7.41 m/s.

(b). It is moving faster at B.

v=ir = 3x15 =45 v