# Particle motion + electric fields

• quicksilver123
In summary, when calculating the final velocity, you should use both the horizontal and vertical displacement values. You should also be careful with the use of angles in the sum of the components, as they are at 90° to each other.
quicksilver123

## Homework Statement

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

vx=4.0*10^6m/s
ε=4.0*10^2 N/C
Δdy=0.02m

mass of electron = me = 9.11*10^-31kg
charge of electron = qe = -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)
Δdy=vi yΔt+0.5(ay)Δt^2
Δt = 2.4*10^-8 seconds

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

c)
vf y^2 = vi y^2 +2ayΔdy
vf y = √(0+1.34884742*10^13)
vf 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.

#### Attachments

• Untitled.jpg
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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.

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?​

quicksilver123 said:
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.​

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?

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

## 1. What is particle motion in relation to electric fields?

Particle motion refers to the movement of charged particles, such as electrons, in response to an electric field. Electric fields are created by the presence of electric charges and can exert a force on other charged particles, causing them to move in a particular direction.

## 2. How are electric fields and particle motion related?

Electric fields and particle motion are closely related because the presence of an electric field can cause charged particles to experience a force and move in a particular direction. The strength and direction of the electric field will determine the motion of the particles.

## 3. What factors affect the motion of particles in an electric field?

The motion of particles in an electric field can be affected by several factors, including the strength and direction of the electric field, the charge and mass of the particles, and any other forces acting on the particles.

## 4. How can electric fields be used to control particle motion?

Electric fields can be used to control particle motion in a variety of ways, such as in particle accelerators or in electronic devices. By manipulating the strength and direction of the electric field, scientists can control the path and speed of charged particles.

## 5. What are some applications of understanding particle motion and electric fields?

Understanding particle motion and electric fields has many practical applications, such as in designing electronic devices, studying the behavior of particles in space, and developing medical imaging techniques. It also plays a crucial role in fields such as physics, chemistry, and engineering.

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