A parallel plate capacitor conists of two flat metal plates placed parallel to e

Question

A parallel plate capacitor conists 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.

a) At what distance from the positively charged plate will the 2 particles pass each other? (Ignore the force of attraction between the 2 particles.)

b) If the electric field were half as much, how would the distance found in part a) change?

Explanation / Answer

using ma = E q

accleration a = E q /m

DIstance travvelled S = 0.5 at^2

distance travelled by proton

dp = 1/2 (eE/mp) t^2

distance travelled by electron

de = 1/2 (eE/me) t^2

pass each other take the same time

dp*mp = de*me

mp = mass of proton

me = mass of electron

mass of proton is 1836 times greater than that of electron.

So the velocity would be in the ratio of 1 : 1836 for proton and electron.

x/(4.20 -x ) = 1/1836

1836x = 4.20 - x

1837 x = 4.20

x = 4.20/1837

x = 2.29*10^-3 cm

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if now E is reduced by 2.

x remains same as x does not depend on Electric field

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