The figure below shows a proton entering a parallel-plate capacitor with a speed

Question

The figure below shows a proton entering a parallel-plate capacitor with a speed of 1.70×105 m/s. The proton travels a horizontal distance x = 7.70 cm through the essentially uniform electric field. The electric field of the capacitor has deflected the proton downward by a distance of d = 0.950 cm at the point where the proton exits the capacitor. (You can neglect the effects of gravity.)

Using kinematics, find the vertical acceleration (including sign) of the proton in this electric field.
Find the magnitude of the force on the proton.
Find the strength of the electric field within the capacitor.
Find the speed of the proton when it exits the capacitor.

Explanation / Answer

Let us start with:-
F=qE
and
F=ma (Newton's second law)

Part A:
Equating gives: ma=qE
a =qE/m

q and m are constants so the ''a' vector is the same as the E vector to be balanced both sides of the equ with the correct sign.
E direction can be easily determined by placing a positive test charge between the plates. The direction in which the positive test charge moves will be the direction of the E field. Since you are using a proton anyway that has a positive charge and the proton is deflected downwards then the E field must point downwards and this is also the direction of the acceleration, that is with the field and positive. This also implies that the top plate is connected to the + volts of the supply and the bottom plate is connected to the negative of the supply. E is from + to -

To determine the value of a you need to solve the following.

s, the deflection on leaving the plates(your 0.95 cm) is given by : s = ut + 1/2 at^2. I am assuming that the proton enters between the plates horizontally on your diagram diagram in which case the u part will be zero, since it has no initial vertical velocity.

s= 0+ 1/2 x qE/m x t^2

t, the time the proton is between the plates and under the influence of the electric field is given by
horizontal distance traveled (d)/velocity (v) note small v
Where: horizontal distance traveled is 7.70 cm or 7.7 x 10^-2 m and v = 1.70 x 10^5 m/s

s =qE/2m x (d/v)^2
E=(2msv^2)/(q x d^2)
E = (2 x 1.67 x 10^-27 x 0.95 x 10^-2 x (1.7 x 10^5)^2)/ (1.6 x 10^-19 x (7.7 x 10^-2)^2)
E = 1.8996 x 10^-18/5.1984 x 10^-22
E = 966.64 N/C
E = 9.67 x 10^2 N/C

(B) F = qE
F = 1.6 x 10^-19 x 9.67 x 10^2
F= 1.547 x 10^-16 N

(C) = 9.67x10^2 N/C as previously calculated

(D)
The proton follows a parabola as in projectile motion
The horizontal component of velocity remains 1.70 x 10^5 m/s throughout flight because no accelerating force applies in horizontal direction. However, although the proton's initially vertical velocity, uy, ist zero at the start, the proton is now moving with vertical velocity, vy on leaving the plates because it is under the influence of a we calculated earlier.

vy^2 = uy^2 + 2 x qE/m x s
But uy = zero
vy^2 =2qEs/m
vy^2 = (2x1.6x10^-19x9.67x10^2x7.7x10^-2)/(1.67x10^-27) = 142.6759 x 10^8 (m/s)^2

The velocity on leaving the plates will be the vector sum of the vertical and horizontal velocities
v = sqr root[vy^2 + vx^2]
v = root of [ 142.6759 x 10^8 + (1.7 x 10^5)2]
v= root of [142.6759 x 10^8 + 2.89 x 10^10]
v =root of [4.316759 x 10^10]
v = 2.078 x 10^5 m/s

So, speed = 2.08 x 10^5 m/s

Now, answer of first part,

Vertical acceleration of the proton, a = qE/m = (1.6x10^-19x9.67x10^2)/(1.67x10^-27) = 9.26 x 10^10 m/s^2

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