A small particle has charge -2.00 μC and mass 2.60×10−4 kg . It moves from point A, where the electric potential is VA = 270 V , to point B, where the electric potential VB = 840 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. What is its speed at point B?
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Charge = q = -2.00*10^-6 C
Mass = m = 2.60×10−4kg
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*2.00*10^-6 (270 -840) / 2.60×10−4 ]
v = sq rt [ 9.61 +4.00*10^-6 (570 ) / 2.60×10−4 ]
v = sq rt [ 9.61 +4.00*10^-2 (285 ) ]
v = sq rt [ 9.61 + 11.4 ]
v = sq rt [ 21.04 ]
v = 4.58 m/s
A) Its speed at point B is 4.58 m/s
B)It is moving faster at B than at A
A small particle has charge -2.00 μC and mass 2.60×10−4 kg . It moves from point...
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