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Speed of Charged Particle

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!

Since there is no other force applied to the particle but only electric force, the total energy must conserved (total energy at point A equals to the total energy at point B).


EA = (1/2)*m*Va^2 + q*VA
EB = (1/2)*m*Vb^2 + q*VB,

=> (1/2)*(1e-4)*(5.6)^2 - (6.9e-6)*280 = (1/2)*(1e-4)*(Vb)^2 - (6.9e-6)*930

=> Vb = 11 (m/s)

The particle is moving faster at B