A small particle has charge `-5.00muC` and mass `2.00 xx 10-4` kg. It moves from point A where the electric potential is `V_A = +200 V`. to point B, where the electric potential is `V_B=+ 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

A, B, D

Explanation

`k_A+U_A=k_B+U_B`
`:. 1/2mv_A^2+qV_A=1/2mv_B^2+qV_B`
`:. v_B=sqrt(v_B^2+2/mq(V_A-V_B)`
`=sqrt((5)^2+(2xx(-5xx10^-6))/(2xx10^-4)(200-800)`
`=7.42m//s`
`v_B-v_A` as the negative charge is moving (freely) from lower potential at `A` to higher potential at `B`. So, its electrostatic `PE` will decrease and kinetic energy will increase.