Physics Experts Need some help with Potential Energy and Potential

[whatisphysics]

Homework Statement

1. How much work is required to move q1 from the negative plate to points A, B, or C?
2. What is the potential energy of q1 at A, B, or C?
3. If q1 is released, what will be the change in its potential energy as it 'falls' to the negative plate from A, B, or C?
4. What will the kinetic energy of q1 when it reaches the negative plate if it is released from A, B, or C?
5. What is the potential(V) at points A, B, and C?

Homework Equations

W = q₀E*h
U = q₀V = k(qq₀/r)
V = U/q₀

The Attempt at a Solution

I am not sure if I am doing these questions right, I just need some hints and someone to kindly point out my mistakes! (:
1. Work is F*d, which is also qE*h; so for A, it would be (1x10^-6C*4x10⁵N/C*0m = 0

2. I am not so certain about this one, would it be the same value as question 1? Because U=q₀E*y?

3. Am I right about this one: the change in potential energy is simply the negative value of question 2?

4. I know that potential energy is converted into kinetic energy, so again, if I'm on the right track, is it just the positive value of question 3?

5. I know that V = U/q₀, but what about the charges? What do I plug into q₀ as? The q1 or the q2?

Thanks to anyone who can kindly shed some light on this! Thank you!

 

[supratim1]

before you proceed, can you tell me the definition of potential difference?

 

[whatisphysics]

before you proceed, can you tell me the definition of potential difference?

According to my textbook, its the difference in charge between two points?

 

[supratim1]

absolutely wrong.

 

[supratim1]

it is - The amount of energy per unit charge needed to move a charged particle from a reference point to a designated point in a static electric field.

 

[Piyu]

I think the formula for work down u may be looking for here is W = V dq

 

[whatisphysics]

it is - The amount of energy per unit charge needed to move a charged particle from a reference point to a designated point in a static electric field.

So, what would the difference in charge between two points be? Is that potential energy then?

thanks for such a quick response! And this might sound silly, what's the difference between potential difference and potential energy? I'm really confused by all the 'potentials'.

 

[tiny-tim]

… what's the difference between potential difference and potential energy? I'm really confused by all the 'potentials'.

hi whatisphysics!

electric potential = potential energy per charge

(same as gravitational potential = potential energy per mass = mgh/m = gh)

So, what would the difference in charge between two points be? Is that potential energy then?

it isn't anything … forget it

 

[whatisphysics]

hi whatisphysics!

electric potential = potential energy per charge

(same as gravitational potential = potential energy per mass = mgh/m = gh)


it isn't anything … forget it

Okay, so for #2, do I use U = qV = k(q/r)? If so, wouldn't that be the exact same value as work? One thing I really don't quite understand is why my textbook says: W = -U

Thanks in advance!

 

[tiny-tim]

potential energy is defined as minus the work done (by a conservative force such as an electric field)

 

[HallsofIvy]

Yes, the difference in potential for two different points is the work required to move an object from one point to another without changing its kinetic energy.

However, "potential" is always relative to some base point. Here, you are told that the reference point is the negative plate. The potential energy of a charge at a point is that energy needed to move the charge from the negative polate to that point.

If a particle is "released" and allowed to "fall" from a point with a higher potential, its kinetic energy is the difference is potential energies.

 

[whatisphysics]

If a particle is "released" and allowed to "fall" from a point with a higher potential, its kinetic energy is the difference is potential energies.

If I'm understanding this correctly, for questions 3 and 4, the difference in potential energy will be equal to the kinetic energy when the charge 'falls' to the negative plate; correct?

So, it would be like: K + U = K + U
Initially at a point, it will have zero kinetic energy, but it has potential energy.
Then, when it falls it loses all the potential energy, and that same value is converted into kinetic energy.

Am I on the right track?

 

[supratim1]

yes, you are on right track. keep going.