Question

Suppose two plates lie in parallel horizontal planes, one plate in the xy plane at  z = 0 and the other plate in the plane that is parallel to the xy plane at z = 10 mm. Between the plates is a constant electric field directed vertically upward (that is, in the positive z direction). A proton and an electron are launched in the positive x direction with the same initial velocity from position (0, 0, 5.0 mm).

Part A :If the proton strikes a plate at (190 mm , 0, 10 mm), where does the electron strike?

Answer 1

electric force=charge*electric field

so the electric force on the proton will be in the vertically upward direction and electric force on the electron is towards vertically downwards.

for the proton:

let the velocity towards +ve x axis is v m/s

force along vertically upward direction=q*E

acceleration=q*E/mp

where mp=mass of proton

let time taken to strike the plate is t seconds.

then 0.5*acceleration*time^2=5 mm

==> time=sqrt(2*0.005/acceleration)...(1)

horizontaol distance=velocity*time

so the horizontal distance , where the proton strikes depends upon time taken to reach the top plate.

time taken varies inversely with square root of acceleration.

acceleration varies inversely with mass.

hence time taken varies directly with square root of mass.

so horizontal distance covered by proton/distance covered by electron=square root of (mass of proton/mass of electron)

=41.931

==> distance covered by electron=190/41.931=4.5312 mm