A small particle has charge -4.60 µC and mass1.70x10-4 kg. It moves from point A, where theelectric potential is VA = 300 V, to point B, wherethe electric potential VB = 820 V is greater thanthe potential at point A. The electric force is theonly force acting on the particle. The particle has aspeed of 6.00 m/s at point A.Part AWhat is its speed at point B?Express your answer in meters per second.For related problemsolving tips and strategies, youmay want to view a Video Tutor Solution of Electricforce and electric potential.m/sPart BIs it moving faster or slower at B than at A?FasterSlower圓

Using the law of conservation of energy we have, 12mvA2+qVA=12mvB2+qVB 12mvB2=q(VA-VB)+12mvA2…

Answered: A small particle has charge -4.60 µC…

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