In the figure, a uniform, upward-pointing electric field E of magnitude 4.50×103
ID: 1524751 • Letter: I
Question
In the figure, a uniform, upward-pointing electric field E of magnitude 4.50×103 N/C has been set up between two horizontal plates by charging the lower plate positively and the upper plate negatively. The plates have length L = 4 cm and separation d = 2.00 cm. Electrons are shot between the plates from the left edge of the lower plate. The first electron has the initial velocity v0, which makes an angle =45° with the lower plate and has a magnitude of 8.93×106 m/s. Will this electron strike one of the plates? If so, what is the horizontal distance from the left edge? If not enter the vertical position at which the particle leaves the space between the plates. Another electron has an initial velocity which has the angle =45° with the lower plate and has a magnitude of 7.14×106 m/s. Will this electron strike one of the plates? If so, what is the horizontal distance from the left edge? If not enter the vertical position at which the particle leaves the space between the plates.
Explanation / Answer
Apply Newton second law on the electron in the electric field is
F = Eq
ma= Eq
a= Eq/ m
=(4.5 * 10^3) (1.6 * 10^-19)/9.11 * 10^-31
=7.9 * 10^14 m/s^2
Appy kinematic equation
vo sin theta - at = 0
t = vo sin theta / a
= (8.93 * 10^6 m/s) sin45/7.9 * 10^14 m/s^2
=0.79 * 10^-8 s
y max = vo t sin tehta - 1/2 * at^2
= vo^2 sin^2 theta/a - 1/2 * a ( vo^2 sin^2 thets/a)^2
= 1/2 a ( vo^2 sin^2 tehts/a^2)
= 1/2 * a * ( 0.79 * 10^-8)^2
= 1/2 * 7.9 * 10^14 m/s^2* ( 0.79 * 10^-8)^2
=2.46 * 10^-2 m
since it is greater than 2.0 cm , the electron might hit the upper plate
(b)
vo sin theta = ( 7.14 * 10^6) sin 45= 5.04* 10^6 m/s
2ad = 2 ( 7.9 * 10^14 m/s^2) (0.02) = 3.16 * 10^13 m/s^2
d = vot sin theta + 1/2 * at^2
1/2 * 7.9 * 10^14 m/s^2 t^2 + 5.04* 10^6 m/s-0.02 = 0
solving quaratic equation
t= 3.17 * 10^-9 s
x= vo t cos theta
= ( 7.14 * 10^6( 3.17 * 10^-9 s) (cos 45
=1.6cm