A charged particle of mass 3.40 10-2 kg and charge +2.10 10-5 C moves from equip
ID: 1286515 • Letter: A
Question
A charged particle of mass 3.40 10-2 kg and charge +2.10 10-5 C moves from equipotential surface A, at a potential of 5200 V, to equipotential surface B, at a potential of 8450 V, under the influence of the existing electric field and an applied force. If the particle has a speed of 2.00 m/s on surface A and arrives at surface B with a speed of 8.00 m/s, how much work is done by the outside force in moving the particle from A to B? (This work is positive if energy is gained by the particle from interaction with the force and negative if energy is lost.) i tried every way on other question similar to this but still got the wrong answer
Explanation / Answer
The potential at A is VA = 5200 V
The potential at B is VB = 8450 V
The mass of the particle is m = 3.40*10-2 kg
The charge of the particle is q = 2.1*10-5 C
The speed of the particle at A is vA = 2 m/s
The speed of the particle at B is vB = 8 m/s
The net work done on the particle will be the change in its total energy
The change in potential energy of the particle is ?U = (?V)(q)
?U = (8450 V - 5200 V)(2.1*10-5 C)
?U = 0.06825 J
The change in kinetic energy of the particle is ?K = 0.5*m*(vB2 - vA2 )
?K = 0.5*(3.40*10-2 kg)*[(8 m/s)2 - (2 m/s)2]
?K = 1.02 J
Therefore the net work done by the force on the charge is
W = 0.06825 J + 1.02 J
W = 1.08825 J
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