A parallel plate capacitor consists of two flat metal plates placed parallel to

Question

A parallel plate capacitor consists of two flat metal plates placed parallel to each other. For this problem assume the plates are held fixed in position a distance of d = 10.00 cm apart, with one plate given a positive charge while the other plate is given an equal amount of negative charge. This set up will produce a uniform electric field between the plates. As shown in the diagram, the electric field points from the positive plate towards the negative plate. Assume the strength of the electric field is 420 N/C. A proton is released from rest at the positive plate and accelerates towards the negative plate. At the same time an electron is released from rest at the negative plate and accelerates towards the positive plate. At what distance from the positively charged plate will the 2 particles pass each other? (Ignore the force of attraction between the 2 particles.) If the electric field were half as much, how would the distance found in ) change?

Explanation / Answer

force on charge , Fe = qE

a = Fe /m = qE/m

for electron:

a_e = (1.6 x 10^-19 x 420 ) / (9.109 x 10^-31) = 7.377 x 10^13 m/s^2

for proton:

a_p = (1.6 x 10^-19 x 420 ) / (1.673 x 10^-27) = 4.017 x 10^10 m/s^2

and both are released from rest hence distance traveled is 0 + at^2 /2

a_e t^2 /2 + a_p t^2 /2 = d

t^2 ( 7.377 x 10^13 + 4.017 x 10^10 ) = 2 x 0.10

t = 5.205 x 10^-8 s

d1 = a_p t^2 /2 = (4.017 x 10^10)(5.208 x 10^-8)^2 /2 = 5.433 x 10^-5 m or 0.0543 mm

b)

E is halved then

a_p' = a_p /2 and a_e' = a_e/2

(a_p' + a_e')t^2 /2 = d

t' = sqrt(2d / (a_p' + a_e'))   = sqrt(2) sqrt(2d / (a_p + a_e))

then distance, d1 = a_p' t'^2 /2 =

        = (a_p/2 ) (sqrt(2) sqrt(2d/ sqrt(2d / (a_p' + a_e')))^2 /2

            = (a_p ) (d / sqrt(d / (a_p' + a_e'))

so distance will not change .

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