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 = 1.85 104 N/C) between two parallel charged plates. The separation of the plates is 1.25 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.

mg/qE is on the order of 10-7 N.

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

mg/qE is on the order of 107 N.

The gravitational force is perpendicular to the electric field.

mg/qE is on the order of 1015 N.

Explanation / Answer

the potential of +ve plate w.r.t. - ve plate V = E*d=1.85*10^4*1.25*10^-2=231.25 volt

the kinetic energy of electron at +ve plate is eV= 3.7*10^-17 J

(1/2)*m*v^2= 3.7*10^-17

v= 9*10^6 m/s

b) electric force Fe =qE= 2.96*10^-15 N

gravitational force Fg=mg=8.93*10^-30 N

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

so we can ignore gravitational force.

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