2
a)
Q = charge = ?
r = distance = 0.46 m
E = dielectric strength = 3 x 106 N/C
Using the equation
E = k Q/r2
3 x 106 = (9 x 109) Q/(0.46)2
Q = 70.5 x 10-6 C
a)
r1 = 50 cm = 0.50 m
electric potential is given as
V1 = k Q/r1 = (9 x 109) (70.5 x 10-6)/(0.50) = 1.3 x 106 Volts
r2 = 80 cm = 0.80 m
electric potential is given as
V2 = k Q/r2 = (9 x 109) (70.5 x 10-6)/(0.80) = 7.93 x 105 Volts
b)
m = mass = 1.76 g = 1.76 x 10-3 kg
q = charge = - 8.4 x 10-9 C
using conservation of energy
qV2 = q V1 + (0.5) m v2
(- 8.4 x 10-9) (7.93 x 105) = (- 8.4 x 10-9) (1.3 x 106 ) + (0.5) (1.76 x 10-3 ) v2
v = 2.2 m/s
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