Oppositely charged parallel plaes are separated by 4.86 mm. A potential differen

Question

Oppositely charged parallel plaes are separated by 4.86 mm. A potential difference of 600 V exists between the plates (a) What is the magnitude of the electric field between the plates? 123457 (b) What is the magnitude of the force on an electron between the plates? Use the value you obtained for the electric field to find the force acting on the electron. N N/C )How much work must be done on the electron to move it to the negative plate if it is initially positioned 2.86 mm from the positive plate? (a) Find the electric potential difference AVe required to stop an electron (called a "stopping potential") moving with an initial speed of 2.42 x 107 m/s. 0,00006889x The response you submitted has the wrong sign. k b) Would a proton traveling at the same speed reqire a reater or lesser magnitude of electric potential difference? greater lesser equal Explain. This answer has not been graded yet (c) Find a symbolic expression for the ratio of the proton stopping potential and the electron stopping potential, Avp/ve (Use the following as necessary: mp for the mass of proton and me for the mass of electron.)

Explanation / Answer

given

d = 4.86 mm = 4.86*10^-3 m
delta_V = 600 V

a) E = deta_V/d

= 600/(4.86*10^-3)

= 123457 N/C

b) magnitude of force on a an electron, |F| = q*E

= 1.6*10^-19*123457

= 1.975*10^-14 N

c) Work must be done = F*d

= 1.975*10^-14*2.86*10^-3

= 6.65*10^-17 J

a) Workdone = change in kinetic energy

q*delta_V = (1/2)*m*v^2

delta_V = m*v^2/(2*q)

= 9.1*10^-31*(2.42*10^7)^2/(2*1.6*10^-19)

= 1665 V

= 1.665 kV

b) greater

becuase proton has more mass than electron.

c) delta_VP/delta_Ve = mp/me

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