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

Question

An electron (mass m = 9.11 times 10^-31 kg) is accelerated in the uniform field E (E = 1.95 times 10^4 N/C) between two parallel charged plates. The separation of the plates is 1.40 cm. The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate, Fig. 16-60. (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 10^15 N. mg/qE is on the order of 10^15 N. mg/qE is on the order of 10^-7 N. The gravitational force is perpendicular to the electric field. mg/qE is on the order of 10^7 N.

Explanation / Answer

BY LAW OF CONSERVATION OF ENERGY---

dEq= 1/2 mv2

THEREFORE V = Sqrt(2dEq/m) = sqrt(2*0.014*19500*1.6e-19/9.11e-31) = 9.8 x 106 m/s answer

gravitational force = mg = 9.11*10-31 * 9.8 N which is negligible as compared to force due to electric field therefore it can be ignored.

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