Electric Potential and Energy in Parallel Plate Configuration

In summary, the electric field between parallel plates is related to the potential energy of an electron. The work needed to move an electron between points I and II is ΔE=V(y-0).
  • #1
skg94
79
0

Homework Statement



Two oppositely charged parallel plates are separated by 0.20m and have an electric potentiial difference of 1.2*10^3V across them. Locations I and II are in the region between the plates. Distance between top and bottom plate is 0.20m.
[itex]\stackrel{}{postive plate\leftrightarrow}[/itex]
| Region 1 : .16m in the y axis, down to region two.
|
|
|
| Region one ends here, Region II is .06m from where region one ends to the point II
[itex]\leftrightarrow[/itex]

I and II create a right angle triangle where 0.16m is the y from I to the right angle, and from the right angle to II is .06m. The voltage is shown to be across the whole two plates.

Homework Equations





The Attempt at a Solution



I really don't have one, i tried V=ΔE/q, but it was just too simple, worth a shot i suppose. I tried finding the difference in voltage didnt work.
 
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  • #2
1. I don't see a question: what are you supposed to find?
2. I don't understand the geometry...

You have a field between two plates, parallel to the x-z plane, with a given PD between them
... from that you can find the electric field between the plates and things like the function V(y).You have a couple of regeons marked out inside the plates ...
from the "top" plate going down - regeon I extends from the top plate down 0.16m
regeon II starts there, and ends 0.06m further down ... at "point II" - which must be below the bottom plate since point II is 0.22m down from the top plate and the plates are 0.2m apart.

Then you say that "I and II create a right angle triangle" which will be gard since you have defined them as regeons (though II could refer to a point instead) and not line-segments.

Perhaps you should scan the diagram?
 
  • #3
The Question is how much work is required to move an electron from location I to II, sorry the II is .06m to the right of where I ends, creating the right trangle.

I will take a picture, and link/upload the pick in a second.

Sorry.
 
  • #5
3011us7.jpg
 
  • #6
Great - now it's clear!

Consider the point where there is a right-angle in the triangle ... call it point III.

Note: the +Q plate is at the top - so the electric field is pointing downwards - so it is a bit like gravity except that electrons are negatively charged. So which direction does the force point? Is the electric field conservative or non-conservative? What does this mean about the relationship between the path electron takes between the endpoints and work needed to follow that path?

What is the relationship between potential energy and work?

What is the potential energy of an electron at points I, II, and III?
Now you can figure out the work.
 
  • #7
Well force is opposite of the field right, electric field is the constant in parallel plates from what i remember. W= change in energy. But how do i start with the distances and such, or are they red herring information? Do i find the hypotenuse?
 
  • #8
F=qE ... so it is in the same direction as the field for positive charges and opposite for negatve charges.

The distances are not a red herring.
Try answering the other questions...

What is the relationship between the potential and the y position?
Between potential and potential energy?
Between potential energy and work?

You've done this with a uniform gravitational field before.
 
  • #9
Well wouldn't an electron want to move up toward the positive plate?

1. I don't really know, between I and III the electric potential is smaller seeing the shorter distance, but electric field remains constant throughout the parallel plates right.
Well if an electron at I wants to move to the negative plate, it has high potential energy right, or is the opposite? Because work is needed to move an electron
 
  • #10
The linchpin is getting the potential function V(y).
Treat the bottom plate as V(y=0)=0, then the top plate is V(y=Y)=?
How does the voltage vary between the plates?
(Hint: the electric field is related to the slope of the potential function - how?)

From there you can get the potential energy of the electron at each point - the change in potential energy is the work.
 

Related to Electric Potential and Energy in Parallel Plate Configuration

1. What is a parallel plates question?

A parallel plates question is a type of problem commonly encountered in physics and engineering, where two parallel plates with opposite charges are separated by a distance and an electric field is present between them. The goal is often to calculate the strength of the electric field or the force acting on a charge placed in the field.

2. How do you calculate the electric field in a parallel plates question?

The electric field in a parallel plates question can be calculated using the equation E = V/d, where E is the electric field strength, V is the potential difference between the plates, and d is the distance between the plates. This assumes that the plates are large enough to be considered infinite and the electric field is uniform between them.

3. What is the difference between a parallel plates question and a capacitor question?

A parallel plates question and a capacitor question are essentially the same type of problem. A capacitor is a device that stores electrical energy in the form of an electric field between two conductive plates. So, in a parallel plates question, the two plates can be thought of as a capacitor, and the focus is on the electric field between the plates.

4. Can the distance between the parallel plates affect the electric field strength?

Yes, the distance between the parallel plates directly affects the electric field strength. As the distance increases, the electric field strength decreases, and vice versa. This is because the electric field is inversely proportional to the distance between the plates, according to the equation E = V/d.

5. Are there any real-life applications of parallel plates questions?

Yes, parallel plates questions have many real-life applications, such as in the design of capacitors, parallel plate air filters, and particle accelerators. They are also used in the study of electrostatics and electricity, which have many practical applications in technology and everyday life.

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