Need help with the second part. In the figure, a uniform, upward electric field

Question

Need help with the second part.

In the figure, a uniform, upward electric field vector E of magnitude 3.50 times 10^3 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. An electron is then shot between the plates from the left edge of the lower plate. The initial velocity vector v_0 of the electron makes an angle theta=45 degree with the lower plate and has a magnitude o 8.12 times l0^6 m/s. Will the 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. 2.66 times 10^2 m The next electron has an initial velocity which has the same angle theta=45 degree with the lower plate and has a magnitude of 6.19 times 10^6 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

1)

along horizantal

initial velocity vox = vo*costheta

ax = 0

from equaiton of motion

x-Xo = vox*T+ 0.5*ax*T^2

x-X0 = vo*costheta*T

T = (x-X0)/(vo*costheta)......(1)

along vertical

voy = vo*sintheta

acceleration ay = -Eq/m = (3.5*10^3*1.6*10^-19)/(9.11*10^-31) = -6.15*10^14 m/s^2

from equation of motion

y-y0 = voy*T + 0.5*ay*T^2

y-y0 = (vo*sintheta*(x-x0))/(vo*costhetra) + (0.5*ay*(x-x0)^2)/(vo^2*(costheta)^2)

y-y0 = (x-X0)*tantheta + ((0.5*ay*(x-X0)^2)/(vo^2*(costheta)^2))

if electron strikes the upper plate

then y - y0 = 0.02 m

0.02 = (x-x0)*tan45 - (0.5*6.15*10^14*(x-x0)^2/(6.15*10^6*cos45)^2)

this has no soltion .so, the electron will not strike the upper plate

the electron will leave the space at the vertical position y

then (x-x0) = L = 4 cm = 0.04 m

y = (0.04*tan45) - (0.5*6.15*10^14*0.04^2/(6.15*10^6*cos45)^2)

y = 0.014 m = 1.4*10^-2 m        <<<<<<===========ANSWER

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