Question

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.

Answer 1

Answer:

(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 $m=2.00\times10^{-4}\ kg$

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

$V_{A}q+\dfrac{1}{2}mv^2=\dfrac{1}{2}mv'^{2}+V_{B}q$

$\dfrac{1}{2}mv'^2=\dfrac{1}{2}mv^2+(V_{A}-V_{B})q$

Put the value into the formula

$\dfrac{1}{2}mv'^2=\dfrac{1}{2}\times2.00\times10^{-4}\times(5.00)^2+(200-800)\times(-5\times10^{-6})$

$\dfrac{1}{2}mv'^2=0.0055$

$v'=\sqrt{\dfrac{2\times0.0055}{2.00\times10^{-4}}}$

$v'=7.41\ m/s$

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

(b). It is moving faster at B.