Question

Any one good with electric fields?

you have a uniform electric field in a given direction, and a given strength. At time o, a particle with a given mass and charge is at the origin, moving with a velocity v at a given angle above the x axis.
1) what is the force exerted on the particle?
2) what are the x and y coordinates of the particle as a function of time?

I really could use some one to explain how to set up the equations, since I am totally lost!

Answer 1

F=Eq, E=V/d, d=s*cos(theta), combine all these

Answer 2

wait a sec!

Answer 3

Okay, what are your variables standing for?

Answer 4

d is the distance between two parallel plates
so I am not sure if it can be used

Answer 5

Yeah, the equations I was finding were for that too, so I'm not sure if I can use those.

Answer 6

That confuses me

Answer 7

I think i am as confused as you are

Answer 8

My professor's notes are no help, and I can't find anything online yet, so I'm stuck.

Answer 9

Thanks for trying at least.

Answer 10

to get da force it is simply u need to apply da equation F=E/Q... about da x and y coordinates u knw da electric field therefore u knw da expression \[E=Q \div4*\pi*\epsilon*R\]. remember their is also an unit vector R multiplied in da numerator.
this unit vector will help u get da coordinates as dis vector is defined with its intial point at da origin... hope i m correct. if i m wrng plz correct me

Answer 11

Hi there - firstly, put formulae for capacitors or coulomb's law out of your head, we don't need them.

The question says there is a uniform electric field, and a particle with charge q.
Well, a particle with charge q experiences a force qE in an electric field, that is something worth remembering. That answers part (1)

We also know that the particle has a mass m, so in response to the force from the electric field, the particle will experience an acceleration given by F/m, or in our case qE/m.

Remember the question said the field is uniform, and so the force and therefore the acceleration of the particle will be uniform.

Okay, we were also told that the particle starts out from the origin with a velocity V.

Now we want to consider the x and y motions separately. Let's say that the electric field has components Ex and Ey, and lets say the particle velocity has components Vx and Vy.

Now we can make use of the equation for displacement for motion under uniform acceleration in a straight line, s(t) = s(0) + ut + at^2/2 and apply it to the x and y directions separately.

For the x direction the acceleration is just qEx/m and since the particle starts out at the origin, we have x(t) = Vx(0)t + qExt^2/2m.

Of course you can write down a similar equation for the y coordinate, y(t) = Vy(0)t + qEyt^2/2m.

Hope that is a bit clearer now.

Answer 12

Thank you!!! I actually understand it now. : )