Motion of a particle in eletric field

• Taylor_1989
In summary, the conversation discusses a question regarding a particle's motion and the use of equations to solve for its vertical position. The conversation covers the use of Newton's second law to identify the net force in the vertical direction and the application of the equation for height to find the particle's position. The final equation obtained is y=(1/2)(E_yq-g)/m)(d^2/v_x^2).
Taylor_1989

Homework Statement

Hi guys. Just wondering if anyone could help me with Q7 part a) on the pdf I have attached below. I think my working are not correct but I can't see other way. If I add 6. and 7. together I would get an ans in terms of what the question is asking but I am not sure if that is correct?

[ note: cropped image inserted by moderator ]

Homework Equations

1.$$y=y_0+v_{0y}(t)+\frac{1}{2}at^2$$
2.$$E_y=\frac{F_y}{q}$$
3. $$F_y=-mg$$

The Attempt at a Solution

4.$$y_0=0, v_y=0 ,\uparrow=+, v_{x0}$$
5.$$E_y=\frac{-mg}{q}, g=\frac{E_yq}{m}$$

So by subbing in the 4 and 5 values into 1 I get two equations as shown below:
6.$$y=-\frac{1}{2}g(\frac{d}{v_{x0}})^2$$
and

7.
$$y=-\frac{1}{2}(\frac{E_yq}{m})(\frac{d}{v_{x0}})^2$$ [/B]

Attachments

• ph322_problems.pdf
621.4 KB · Views: 181
#3 only follows from #2 in a specific situation - does that situation apply for 7a?

You should start from a free body diagram.
Use: ##\sum \vec F = m\vec a## ... divide into components and use suvat equations. Y
ou've done something similar when you learned about ballistics.

Simon Bridge said:
#3 only follows from #2 in a specific situation - does that situation apply for 7a?

You should start from a free body diagram.
Use: ##\sum \vec F = m\vec a## ... divide into components and use suvat equations. Y
ou've done something similar when you learned about ballistics.

I am slight lost to what you mean. I have used suvat, in the x and y. Could you please expand on this. From the picture the only accleration acting is in the horizontal no acceleration in the vertical. So the horizontal equation looks like so:

$$d=v_xt$$

Last edited:
Taylor_1989 said:
the only accleration acting is in the horizontal no acceleration in the vertical
You meant the other way around, yes?
What forces act in the vertical? What is the net vertical force? What vertical acceleration results?

Taylor_1989 said:
I am slight lost to what you mean. I have used suvat, in the x and y. Could you please expand on this.
Identify the vertical forces and write out Newton's second law accordingly.
From the picture the only acceleration acting is in the [vertical] no acceleration in the [horizontal]. So the horizontal equation looks like so: $$d=v_xt$$
So, in the horizontal direction, Newton's law says: ##\sum F_x = 0 = ma_x: v_x(0)=u, x(0)=0##
... when you solve that, you get ##a_x=0, v_x(t)=u, \implies x(t)=ut##

Good ... so what are the forces in the vertical direction?
##\sum F_y = ? = ma_y : v_y(0)=?, y(0)=?## ... fill in the question marks.

Simon Bridge said:
Identify the vertical forces and write out Newton's second law accordingly.
So, in the horizontal direction, Newton's law says: ##\sum F_x = 0 = ma_x: v_x(0)=u, x(0)=0##
... when you solve that, you get ##a_x=0, v_x(t)=u, \implies x(t)=ut##

Good ... so what are the forces in the vertical direction?
##\sum F_y = ? = ma_y : v_y(0)=?, y(0)=?## ... fill in the question marks.
Ok so the net force that act of the particle is:
$$F_t=E_yq-g=ma_y$$

using the equation for height:

$$y=v_{y0}t+\frac{1}{2}a_yt^2$$

now at $$t=0, v_{y0}=0$$

therfore my equation becomes:

$$y=\frac{1}{2}at^2$$

subbing in my equations for $$a_y ,t$$

I get:

$$y=(\frac{1}{2})(\frac{E_yq-g}{m})(\frac{d^2}{v_x^2})$$

Taylor_1989 said:
Ok so the net force that act of the particle is:
$$F_t=E_yq-g=ma_y$$

using the equation for height:

$$y=v_{y0}t+\frac{1}{2}a_yt^2$$

now at $$t=0, v_{y0}=0$$

therfore my equation becomes:

$$y=\frac{1}{2}at^2$$

subbing in my equations for $$a_y ,t$$

I get:

$$y=(\frac{1}{2})(\frac{E_yq-g}{m})(\frac{d^2}{v_x^2})$$
Looks right.

Just like to say thank for the guidance.

Taylor_1989 said:
Ok so the net force that act of the particle is:
$$F_t=E_yq-g=ma_y$$
Force of gravity is usually written ##mg##.

Note: it is usually best practise to use the variable names in the question.
##d = uT## and ##2h = aT^2## where T is time between plates... maybe?

Simon Bridge said:
Force of gravity is usually written ##mg##.

Note: it is usually best practise to use the variable names in the question.
##d = uT## and ##2h = aT^2## where T is time between plates... maybe?
yes sorry, I notice that when i looked back. But i could not edit it.

1. What is the motion of a particle in an electric field?

The motion of a particle in an electric field is the movement of a charged particle in the presence of an electric field. The particle will experience a force due to the electric field, causing it to accelerate and move in a certain direction.

2. What factors affect the motion of a particle in an electric field?

The motion of a particle in an electric field is affected by the strength of the electric field, the charge of the particle, and the mass of the particle. The direction of the electric field also plays a role in determining the direction of the particle's motion.

3. How does the direction of the electric field affect the motion of a particle?

The direction of the electric field will determine the direction of the force exerted on the particle. If the electric field is in the same direction as the particle's initial velocity, it will accelerate the particle in that direction. If the electric field is in the opposite direction, it will decelerate the particle.

4. What is the relationship between the electric field and the particle's velocity?

The electric field and the particle's velocity have a direct relationship. As the strength of the electric field increases, the force on the particle increases, causing it to accelerate and increase its velocity. On the other hand, a decrease in the electric field will result in a decrease in the particle's velocity.

5. How is the motion of a charged particle in an electric field different from that of a neutral particle?

A charged particle in an electric field will experience a force and accelerate, while a neutral particle will not be affected by the electric field. This is because neutral particles have an equal number of positive and negative charges, resulting in a net charge of zero and no interaction with the electric field.

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