# Electric Field Strength Between Two Parallel Plates Problem

• liamporter1702
In summary, the electric field would use 1.49x10^-19 of energy to move an electron 3mm nearer to the positively charged plate.
liamporter1702

## Homework Statement

If the electric field strength between two parallel plates 1cm apart is 3.1kN/C, how much energy will be used by the electric field to move an electron (placed half way between the plates) 3mm nearer to the positively charged plate?

## Homework Equations

Work Done = 1/2V^2C
F=KxQ1Q2/r^2
E=KxQ/r^2
E=V/D

## The Attempt at a Solution

I know that the answer to this question is 1.49x10^-18 J. I feel that I may be going about this question the complete wrong way but this is what I did:

I worked out the potential difference between the plates to be 31V by using the equation V=ED
V=(3.1x10^3)x(0.2x10^-2)-----> I got this value for D by taking away 3mm from the 5mm half way point between the plates.

I then subbed in the values that I know into the equation Work Done = 1/2V^2C
Work Done = 1/2 x 6.2^2 x (1.6x10^-19)
This gave me 3.08x10^-18.

I feel like I'm going the wrong way about this question. Hope someone can help and hope I've worded this well enough for you all to understand haha.

liamporter1702 said:
Yes you are. What's the most basic definition of work? What's the force acting on the electron?

Doc Al said:
Yes you are. What's the most basic definition of work? What's the force acting on the electron?

Work Done = Force x Distance? The force produced by the electric field?
Should I use F = KxQ/r^2 to find the force and then substitute my answer into the Work Done = FxD equation?

liamporter1702 said:
Work Done = Force x Distance? The force produced by the electric field?
Right and right.

Should I use F = KxQ/r^2 to find the force and then substitute my answer into the Work Done = FxD equation?
No, it's easier than that. You are given the field strength! (It's a uniform field between those plates, not the field from a point charge.)

1 person
Doc Al said:
No, it's easier than that. You are given the field strength! (It's a uniform field between those plates, not the field from a point charge.)

Ah yes! I see where you're coming from! So if I have the the field strength of 3.1x10^3 and use the equation, Force = Electric Field Strength x Charge, to work out the force this gives me:
3.1x10^3 x 1.6x10^-19 = 4.96x10^-16

Then apply this to the work done equation:
4.96x10^-16 x 0.3x10^-2 = 1.49x10^-19

Thank you for your help, I really appreciate it!

## 1. What is the formula for calculating the electric field strength between two parallel plates?

The electric field strength between two parallel plates is determined by the formula 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.

## 2. How does the distance between the plates affect the electric field strength?

The electric field strength is inversely proportional to the distance between the plates. This means that as the distance between the plates increases, the electric field strength decreases, and vice versa.

## 3. What is the direction of the electric field between two parallel plates?

The electric field between two parallel plates is always perpendicular to the plates and parallel to the direction of the potential difference.

## 4. How does the potential difference between the plates affect the electric field strength?

The electric field strength is directly proportional to the potential difference between the plates. This means that as the potential difference increases, the electric field strength also increases.

## 5. Can the electric field strength between two parallel plates ever be zero?

Yes, the electric field strength between two parallel plates can be zero if the potential difference between the plates is also zero. This can happen if the plates are connected to the same potential source or if the plates are infinitely large and the distance between them is infinitely small.

• Introductory Physics Homework Help
Replies
58
Views
4K
• Introductory Physics Homework Help
Replies
6
Views
546
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
14
Views
940
• Introductory Physics Homework Help
Replies
2
Views
979
• Introductory Physics Homework Help
Replies
26
Views
876
• Introductory Physics Homework Help
Replies
10
Views
4K
• Introductory Physics Homework Help
Replies
4
Views
4K
• Introductory Physics Homework Help
Replies
3
Views
1K