Question

A 2.3 mm -diameter sphere is charged to -4.5 nC . An electron fired directly at the sphere from far away comes to within 0.33 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?

Answer 1

R = radius of sphere = diameter/2 = 2.3/2 = 1.15 mm = 1.15 x 10^-3 m

Q = charge on the sphere = - 4.5 x 10^-9C

q = charge on electron = - 1.6 x 10^-19 C

v = initial speed = ?

m = mass of electron = 9.1 x 10^-31 kg

r = closest distance of electron from center of sphere = R + 0.33 = 1.15 + 0.33 = 1.48 mm = 0.00148 m

using conservation of energy

kinetic energy = final electric potential energy

(0.5) m v² = k Q q/r

(0.5) (9.1 x 10^-31) v² = (9 x 10⁹) (- 4.5 x 10^-9) (- 1.6 x 10^-19)/(0.00148)

v = 9.81 x 10⁷ m/s

b)

v_i = initial speed = v = 9.81 x 10⁷ m/s

v_f = final speed = v_i/2

r_f = final distance between electron and center of sphere

Using conservation of energy

initial KE = final KE + final electric potential energy

(0.5) m v_i² = (0.5) m v_f² + k Q q/r_f

(0.5) (9.1 x 10^-31)  (9.81 x 10⁷)² = (0.5) (9.1 x 10^-31)  ((9.81 x 10⁷)/2)² + (9 x 10⁹) (- 4.5 x 10^-9) (- 1.6 x 10^-19)/r_f

r_f = 1.97 x 10^-3 m = 1.97 mm

d = distance from surface = r_f - R = 1.97 - 1.15 = 0.82 mm

c)

r = distance between electron and center at turning point = 1.48 mm = 1.48 x 10^-3 m

F = force on the electron

F = k Qq/r² = (9 x 10⁹) (4.5 x 10^-9) (1.6 x 10^-19)/(1.48 x 10^-3)² = 1.96 x 10^-12 N

m = mass of electron = 9.1 x 10^-31 kg

acceleration is given as

a = F/m = (1.96 x 10^-12)/(9.1 x 10^-31) = 2.2 x 10¹⁸ m/s²