a5. The figure shows a proton (mass - m, charge te) at rest in a vacuum between

Question

a5. The figure shows a proton (mass - m, charge te) at rest in a vacuum between two parallel plates (each plate is square with side length, L- 100.0m) oriented horizontally, one above the other. By setting the potental difference between the plates to AV-2.0458x10 volts, the induced charge on the plates Q, generates an upward vertical electric force on the proton which exactly cancels the downward gravitatinal foass it remains stationary in between the plates). Thus knowing that the charge of a proton is 1.6 the proton can be determined as follows. force on it (thus of a proton is 1.6x101 C, the mass o -Q FE AV-2.0458 x 10-8v proton D 0.2m +Q ? 100.0m ? (a) Calculate the magnitude of the electric field in the space between the plates. (ANS: 1.0229x10" N/Cc) (b) What is the induced charge on the positively-charged plate? (c) Use Newton's first law and the result form Part (a) to show that the mass (m) of the proton is 1.67x10

Explanation / Answer

a)We know that

E = V/d

E = 2.0458 x 10^-8/0.2 = 1.0299 x 10^-7 N/C

Hence, E = 1.0229 x 10^-7 N/C

b)we know that, electric field between the plates is:

E = sigma/e0

sigma = E e0

sigma = 1.0229 x 10^-7 x 8.85 x 10-12 = 9.0527 x 10^-19 C/m^2

sigma = Q/A =>Q = sigma A

Q = 9.0527 x 10^-19 x 100^2 = 9.0527 x 10^-15 C

Hence, Q = 9.0527 x 10^-15 C

c)As stated

Fg = Fe

mg = qE => m = qE/g

m = 1.67 x 10^-19 x 1.0229 x 10^-7/9.81 = 1.67 x 10^-27 kg

Hence, m = 1.67 x 10^-27 kg

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