An electron (mass m = 9.11 10 -31 kg) is accelerated in the uniform field E ( E

Question

An electron (mass m = 9.11 10-31 kg) is accelerated in the uniform field E (E = 1.90 104 N/C) between two parallel charged plates. The separation of the plates is 1.20 cm. The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate.

(a) With what speed does it leave the hole?
m/s
(b) Explain why the gravitational force can be ignored.

The gravitational force is perpendicular to the electric field.mg/qE is on the order of 10-15 N.     mg/qE is on the order of 107 N.mg/qE is on the order of 1015 N.mg/qE is on the order of 10-7 N.

Explanation / Answer

a) Use Work-energy theorem

Gain in kinetic of electron = Workdone on elctron

(1/2)*m*v^2 = F*d

(1/2)*m*v^2 = q*E*d

==> v = sqrt(2*q*E*d/m)

= sqrt(2*1.6*10^-19*1.9*10^4*0.012/(9.11*10^-31))

= 8.95*10^6 m/s

b) mg/qE is on the order of 10-15 N.

here,
m*g/(q*E) = 9.11*10^-31*9.8/(1.6*10^-19*1.9*10^4)

= 2.9*10^-15

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