To find the parameters of velocity in an electric field

[PhysicsTest]

Homework Statement

The essential features of the displaying tube of an oscilloscope are shown. The voltage difference between K and A is Va and between P1 and P2 is Vp. Neither electric field affects the other one. The electrons are emitted from the electrode K with initial 0 velocity, and they pass through a hole in the middle of electrode A. Because of the field between P1 and P2 they change direction while they pass through these plates and, after that, move with constant velocity toward the screen S. The distance between plates is d.
a. Find the velocity vx of the electrons as a function of Va as they cross A.
b. Find the Y- component of velocity vy of the electrons as they come out of the field of plates P1 and P2 as a function of Vp,ld,d,vx
c. Find the distance from the middle of the screen (ds), when the electrons reach the screen, as a function of tube distances and applied voltages.
d. For Va=1.0kV, and Vp =10V, ld=1.27cm, d=0.475 cm and ls=19.4cm, find the numerical values of vx, vy and ds.
e. if we want to have a deflection of ds=10cm of the electron beam, what must be the value of Va?

Relevant Equations

qE =ma; mv^2=qV

i have drawn the E field as below, hence the F will be in the upward direction for electron

a. Using energy is constant, the velocity $v_x$ as it crosses A is
$0.5mv_x^2 = q*V_a$
$v_x = \sqrt{(\frac{2qV_a} m)} m/s$
one doubt i have here is, the question mentions electrons, but i have taken the mass of single electron. Is it correct?
b. $qE = ma$
$\frac{mdv_y} {dt} = \frac {qV_p} d$
$v_y = \frac {qV_pt} {md} + C1$ $v_y=0 at t=0$ Hence $C1=0$
$v_y = \frac {qV_pt} {md}$ -->eq1
$v_x \text{ moves with constant velocity as there is no force in x-direction}$
$t = \frac{l_d} {v_x}$ -->eq2, substitute in eq1
$v_y = \frac{qV_pl_d} {mdv_x} m/s$ --> eq3
I still need to attempt other parts, but am i in the right direction?

 

[kuruman]

It is safe to assume that all electrons follow the same path, therefore a one-electron calculation is sufficient.

You are on the correct path so far. Thank you for using equations in symbolic form; it is appreciated.

 

[kuruman]

By the way, when you write equations in symbolic form, you don't have to supply units. Units must be supplied when you substitute numbers for symbols. For example $t=\frac{l_d}{v_x}$ has dimension of time as noted by the ratio of distance to speed but the units are undefined. If you substitute $l_d=10~\text{m}$ and $v_x=2~\text{m/s}$, then $t = 5~\text{s}$, but if $v_x = 2~\text{m/hr}$ then $t = 5~\text{hours}.$ Same equation, same numbers but different units.

 

[PhysicsTest]

C. Seems to be slightly tough, From eq(3) post #1. The
$v_y=\frac{qV_pl_d} {mdv_x}$
$dy = \frac{qV_pl_ddt} {mdv_x}$
$y = \frac{qV_pl_dt} {mdv_x}$ --> eq(4), substitute "t" from eq(2) in eq(4),
$t = \frac{l_d} {v_x}$
$y = \frac{qV_pl_d^2} {mdv_x^2}$ --> eq(5) when electron crosses the plate
Let "t1" be the time it takes to hit the screen, it travels with constant speed on both x and y-axis (constant vx,vy).
To cover a distance of $l_s$ on the x-axis
$v_x = \frac{l_s} {t1}$ --> eq6
$v_y = \frac{d_s-y} {t1}$ --> eq7
$\frac{v_x} {v_y} = \frac{l_s} {d_s-y}$
$v_x(d_s-y) = l_sv_y$
$d_s = \frac{l_sv_y} {v_x} + y$ --> eq(8)
Is the answer correct? I have few doubts on the eq(7), do i need to the subtraction.

 

[kuruman]

Not correct. From post #1 Eq. 1 you have $v_y = \frac {qV_pt} {md}$ which makes $dy = \frac{qV_pl_ddt} {mdv_x}$ in post #8 incorrect because $dy=v_y dt=\frac {qV_p} {md}t dt.$

Perhaps part (c) will be conceptually easier for you if you note that when the electron accelerates for the second time in a direction perpendicular to its initial velocity, this problem is equivalent to throwing a rock initially in the horizontal direction from some height above ground. The only difference is that the acceleration here is $a=\frac{qV_p}{md}$ instead of $g$.

 

[PhysicsTest]

Sorry my mistake, I am not very much thorough with kinematics as well I have not attempted many problems, I know basic kinematic equations. I will attempt the (C) problem with basics i know.

Step1: To calculate the distance 'y' under the influence of the plate voltage, it is
$dy = v_y dt = \frac{qV_p}{md} tdt$
$y = \frac{qV_pt^2} {2md} +C1 \text{ t=0 y=0; C1=0}$ ---> eq4
As per eq2 post 1 $t = \frac{l_d} {v_x}$ substitute in eq4
$y = \frac{qV_pl_d^2}{2mdv_x^2}$ This is the y-distance e- will travel under force
After the plate it will start moving with constant velocity in $v_x$ and $v_y$direction
Let it take "t1" time to travel x distance = $l_s$ (to touch screen) with speed $v_x$
In the same time "t1" time to travel y distance = $d_s$ with speed $v_y$
$\frac{l_s} {v_x} = \frac{d_s-y} {v_y}$
$d_s = y + \frac{v_yl_s} {v_x}$ ---> eq5 ($v_y$ is derived in post #1 eq3 )
I hope i have done correctly? I am treating the $v_x, v_y$ as independent and solved. Is it correct?

 

[kuruman]

Well done! The expression $d_s = y + \frac{v_yl_s} {v_x}$ is correct. However, in part (c) you have to find $d_s$ "as a function of tube distances and applied voltages" which means one more substitution for $y$, $v_x$ and $v_y$.