27 Jun, 2021 06:57 PM

A 2.6 mm-diameter sphere is charged to -4.6 nC . An electron fired directly at the sphere?

from far away comes to within 0.39 mm of the surface of the target before being reflected. a) What was the electron's initial speed? b) At what distance from the surface of the sphere is the electron's speed half of its initial value? c) What is the acceleration of the electron at its turning point? Thank you

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27 Jun, 2021 06:57 PM


electron's initial kinetic energy = electron's electric potential energy before being reflected

mv^2/2 = (q)(e)/[(4πεo)(b)]

(9.11e-31 kg)(v^2)/2 = (-4.6e-9 C)(-1.602e-19 C)/[(4π)(8.85e-12 C^2/N-m^2)(2.6e-3 / 2 + 3.9e-4 m)]

v = 9.3e7 m/s


electron's initial kinetic energy = (kinetic energy of electron) + (electric potential energy of electron)

mv^2/2 = m(v/2)^2/2 + (q)(e)/[(4πεo)(r)]

3mv^2/8 = (q)(e)/[(4πεo)(r)]

3(9.11e-31 kg)(9.3e7 m/s)^2 / 8 = (-4.6e-9 C)(-1.602e-19 C)/[(4π)(8.85e-12 C^2/N-m^2)(2.6e-3 / 2 m + x)]

x = 9.5e-4 m = 0.95 mm


ma = (q)(e)/[(4πεo)(b)^2]

(9.11e-31 kg)(a) = (-4.6e-9 C)(-1.602e-19 C)/[(4π)(8.85e-12 C^2/N-m^2)(2.6e-3 / 2 + 3.9e-4 m)^2]

a = 2.5e18 m/s^2

27 Jun, 2021 06:57 PM

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