For neatness and convenience, I'll use the notation 1.6E-19 (for example) instead of 1.6x10⁻¹⁹.
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Proton's initial kinetic energy K1 = ½mv₀²
= ½ x 1.67E-27 x (1.6E6)²
= 2.138E-15J
The potential energy at the start point P1 = kqQ/r
= 9E9 x 1.6E-19 x -10E-9/0.003
= -4.8E-15 J
The potential energy at the end point P2 = kqQ/r
= 9E9 x 1.6E-19 x -10E-9/0.004
= -3.6E-15 J
Since energy is conserved:
K1 P1 = K2 P2
K2 = K1 P1 - P2
. . .= 2.138E-15 (-4.8E-15) - (-3.6E-15)
. . .= 9.38E-16 J
½mv² = 9.38E-16
v = √[2 x 9.38E-16 / 1.67E-27]
. .= 1.06E6 m/s
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Of course the solution can be written much quicker than that. I would usually just write:
½mv₀² kqQ/r1 = ½mv² kqQ/r2
½mv² = ½mv₀² kqQ(1/r1 - 1/r2)
and then substitute values.
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