Particle motion + electric fields

[quicksilver123]

Homework Statement

see attachment for question wording.
a) find acceleration
b) find horizontal displacement
c) find final velocity

v_x=4.0*10^6m/s
ε=4.0*10^2 N/C
Δd_y=0.02m

mass of electron = m_e = 9.11*10^-31kg
charge of electron = q_e = -1.6*10^-19 C

Homework Equations

suvat
electric field equation
coulombs laws

The Attempt at a Solution

a)
ma=qε
(9.11*10^-31)(a)=(1.6*10^-19)(4*10^2)
acceleration = 7.025246981*10^13 m/s/s [down]

b)
Δd_y=v_{i y}Δt+0.5(a_y)Δt^2
Δt = 2.4*10^-8 seconds

Δd_x = vΔt
Δd_x = (4*10^6)*2.4*10^-8)
Δd_x = 0.096m
Δd_x = 9.6cm

c)
v_{f y}^2 = v_{i y}^2 +2a_yΔd_y
v_{f y} = √(0+1.34884742*10^13)
v_{f y} = 3.672665817*10^6 m/s



I'm pretty confident in parts A and B but I'm not sure about part C.
I still have to find the direction of the final velocity; waiting to see if the magnitude is correct first.

 

[gneill]

Your methods look okay. The results for a and b look good, but watch your significant figures.

Something went wrong in the numerical calculation for part c; the value you obtained for the vertical velocity is not correct. Also, presumably the final velocity should include both the horizontal and vertical components.

 

[quicksilver123]

Ah. I used the horizontal displacement found in part b instead of the vertical displacement like I should've.

Let

and [down] be positive.

c)
vf y^2 = vi y^2 +2ayΔdy
vf y = √(0+2((7.025246981)(0.02))
vf y = 0.167633493 m/s [down]

vf x = √(vi x^2 + 2(0)(0.096)
vf x = vi x
vf x = 4*10^6 m/s

theta = tan^-1((4*10^6)/(0.167633493))
theta = 89.9999976°
so, I am thinking this is negligible.


vf = √(0.167633493^2 + (4*10^6)^2 )
vf = 2000.00007 m/s

is this correct?​

 

[gneill]

Ah. I used the horizontal displacement found in part b instead of the vertical displacement like I should've.

Let

and [down] be positive.

c)
vf y^2 = vi y^2 +2ayΔdy
vf y = √(0+2((7.025246981)(0.02))
vf y = 0.167633493 m/s [down]
​

That seems way to small for the y-velocity. Check the acceleration value you've used.

vf x = √(vi x^2 + 2(0)(0.096)
vf x = vi x
vf x = 4*10^6 m/s

In other words the x velocity is unchanged because there is no acceleration in the x direction.

theta = tan^-1((4*10^6)/(0.167633493))
theta = 89.9999976°
so, I am thinking this is negligible.

Don't use angles to calculate the sum of the components, they are at 90° to each other. Use Pythagoras: square root of sum of squares.

vf = √(0.167633493^2 + (4*10^6)^2 )
vf = 2000.00007 m/s

is this correct?​

Nope. Fix the bits above and recalculate.​

 

[quicksilver123]

Ugh. Careless mistake. Seems I'm making a lot of those... :/

Fixed:


vf y^2 = vi y^2 +2ayΔdy
vf y = √(0+2(7.025246981*10^13)(0.02)
vf y = √(2.80098792*10^12)
vf y = 1.676334928*10^6 m/s

vf x = vi x = 4*10^6 m/s

theta = tan^-1 (vfx/vfy) = 67.26 degrees

vf = √((1.676334928*10^6)^2+(4*10^6)^2)
vf = 4.337061078*10^5 m/s [down 67.26 degrees right]


finally.

look good?

 

[gneill]

Much better. Be sure to round submitted results to appropriate significant figures.

 

[quicksilver123]

Thanks!

I also have some questions here:

https://www.physicsforums.com/showthread.php?t=667813

and here

https://www.physicsforums.com/showthread.php?t=580224

if you don't mind :)