A small particle has charge 2.90uC and mass 2.00×10−4kg . It moves from point A, where the electric potential is VA= 200 V, to point B, where the electric potential VB= 860 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 3.10 m/s at point A.
A) What is its speed at point B ?
B) Is it moving faster or slower at B than at A?
Faster or Slower
1 Answer

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Charge = q = 2.90*10^6 C
Mass = m = 2.00×10−4kg
Initial velocity = u =3.10 m/s
Final velocity = y
Change in kinetic energy = change in potential energy
(1/2)mv^2 – (1/2)mu^2 = q (Va – Vb)
multiply by 2 abd divide by m on both sides
v^2 – u^2 = 2q (Va – Vb) /m
v^2 = u^2 + 2q (Vb – Va) /m
v = sq rt [ u^2 + 2q (Va – Vb) /m ]
v = sq rt [ 3.1^2 – 2*2.90*10^6 (200 860) / 2.00×10−4 ]
v = sq rt [ 9.61 + 5.80*10^6 (660 ) / 2.00×10−4 ]
v = sq rt [ 9.61 + 5.80*10^2 (330 ) ]
v = sq rt [ 9.61 + 19.14 ]
v = sq rt [ 28.75 ]
v = 5.361 m/s
A) Its speed at point B is 5.361 m/s
B)It is moving faster at B than at A
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