A small particle has charge -6.90 micro-Coulombs and mass 1.00×10-4 kg . It moves from point A, where the electric potential is VA= 280 V , to point B, where the electric potential is VB= 930 V; note that this 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 5.60 m/s at point A.
What is its speed at point B?
Is it moving faster or slower at B than at A?
Solution This solution is FREE courtesy of BrainMass!
EA = EB
EA = (1/2)*m*Va^2 + q*VA
EB = (1/2)*m*Vb^2 + q*VB,
=> Vb = 11 (m/s)
The particle is moving faster at B