Two identical charges 2 80 Meu C are fixed on the y axis at

Two identical charges (2 = 8.0 Meu C) are fixed on the y axis at y = plusminus 0.30 m {see Fig. I A particle that has m = 4.0 X 10^-4 kg and q = 5.0 Meu C has a velocity v = vi when its position is r = -0.40i m. Calculate the minimum value of v for which the particle will reach x = 0. If v = 40 m/s, how close will the particle get to x - 0? If v = 100 m/s, how fast will the particle be moving as it passes through x = 0?

Solution

Lets find the Potential enegy at x = 0 and x = -0.4 m

At 0 m

V = 2K Qq/r = 2*9*10^9*8*5*10^-12/.3 = 2.4 J

At 0 m

V = 2K Qq/r = 2*9*10^9*8*5*10^-12/.5 = 1.44 J

Hence we need at least

E = 2.4 - 1.44 =.96 J

Hence

1/2 m V^2 = .96

V = 69.28 ms-1

b) With V = 40ms-1 we reach

1.44 + 1/2 m V^2 = 2K Qq/r = 2*9*10^9*8*5*10^-12/r

r = .27 m

c) Total energy has to be conserved

1.44 + 1/2 m V^2 = 2.4 + 1/2*m *V\'^2

V\' = 72 ms-1