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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, |
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Answer» Correct Answer - A::B::D `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. |
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