A small particle has charge -4.90 μC and mass 1.90*10-4 kg. It moves from point A, where the electric potential is VA = 200 V, to point B, where the electric potential VB = 810 V is greater than the potential at point A. The electric force is the only force acting on the particle. The particle has a speed of 4.70 m/s at point A. For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Electric force and electric potential. Part A What is its speed at point B? Express your answer in meters per second. V= Submit Part B VE ΑΣΦ Request Answer Is it moving faster or slower at B than at A? O Faster O Slower Submit Request Answer ? m/s

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Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.