Change in electric potential between 2 parallel plates

Click For Summary
SUMMARY

The change in electric potential between two parallel plates in a uniform electric field of 90,000 V/m can be calculated using the formula for electric potential difference, which is the product of the electric field strength and the displacement in the direction of the field. For a proton displaced 0.2 m, the change in electric potential is 90,000 V/m multiplied by 0.2 m, resulting in a potential difference of 18,000 volts. The correct approach involves understanding that the work done by the electric field on the charge is equal to the change in electric potential energy.

PREREQUISITES
  • Understanding of electric fields and potential difference
  • Familiarity with the concept of work done by electric fields
  • Knowledge of basic physics equations related to electric potential
  • Ability to manipulate units and perform calculations involving volts and meters
NEXT STEPS
  • Study the relationship between electric field strength and potential difference
  • Learn about the work-energy principle in electric fields
  • Explore examples of electric potential calculations in uniform fields
  • Review physics textbooks or online resources specifically covering electric potential
USEFUL FOR

Students studying physics, particularly those focusing on electromagnetism, as well as educators seeking to clarify concepts related to electric potential and fields.

Runaway
Messages
48
Reaction score
0

Homework Statement


A proton is released from rest in a uniform electric field of mag. 90000 V/m directed along the positive x axis. the proton undergoes a displacement of .2m in the direction of the electric field. Find the change in electric potential.


Homework Equations



no idea

The Attempt at a Solution


I have no idea where to start, an equation would be nice but I can't find anything in the book about electric potential in a uniform electric field.
First I tried 90000V/m * .2m but that didn't work, then I tried using V= kq/r, V=k * <charge of a proton> / .2m, but that didn't work either.
 
Physics news on Phys.org
Electric potential difference is minus the work done by the electric field divided by the charge. So the question is, what work would the electric field do on this charge over that distance?

Any physics textbook worth its salt should explain this, so I suggest looking harder or getting a new book.
 

Similar threads

Two different dielectrics between parallel-plate capacitor
  • · Replies 6 ·
Replies
6
Views
2K
Why does electric potential energy increase if you move against the field?
  • · Replies 6 ·
Replies
6
Views
2K
Earthing of two parallel conducting plates
  • · Replies 11 ·
Replies
11
Views
2K
The sign of the change in potential
  • · Replies 7 ·
Replies
7
Views
1K
Electric Potential Energy of Point Charge Systems
  • · Replies 2 ·
Replies
2
Views
1K
Electric field between two parallel plates
  • · Replies 58 ·
2
Replies
58
Views
6K
Electric fields of charged plates
  • · Replies 6 ·
Replies
6
Views
2K
Electric potential due to shell containing a charge at an offset outside
  • · Replies 5 ·
Replies
5
Views
1K
A disconnected capacitor with two dielectrics in parallel
  • · Replies 8 ·
Replies
8
Views
2K
Parallel Plate Capacitor Filled with Many dielectrics
  • · Replies 20 ·
Replies
20
Views
4K