A small particle has charge -4.10 μCμC and mass 1.80×10−4 kgkg. It moves from point AA, where the electric potential is VAVAV_A = 230 VV, to point BB, where the electric potential VBVBV_B = 980 VV is greater than the potential at point AA. The electric force is the only force acting on the particle. The particle has a speed of 5.20 m/sm/s at point AA. What is its speed at point BB? Express your answer in meters per second.
A small particle has charge -4.10 μCμC and mass 1.80×10−4 kgkg. It moves from point AA, where the electric potential is VAVAV_A = 230 VV, to point BB, where the electric potential VBVBV_B = 980 VV is greater than the potential at point AA. The electric force is the only force acting on the particle. The particle has a speed of 5.20 m/sm/s at point AA.
The given particle has a charge of and a mass of
This particle is initially at point A, where the potential is 230 V, and later at point B, where the potential is 980 V
A charge kept in a potential field has a potential energy given as
At point A, the particle has a speed of 5.2 m/s, and thus owing to this speed the particle will also have a kinetic energy at A
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