Electric potential at an unknown point

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SUMMARY

The discussion focuses on calculating the electric potential at point XB (0 m) within a parallel plate capacitor, given an electric field strength of E = -3500 N/C and a known potential at point XA (3.00 m) of 1500 V. The relationship used is V = E ⋅ s, leading to the calculation of displacement s as 2.333 m. Participants emphasize the importance of correctly interpreting the electric field direction and potential differences, as well as the necessity of sketching the configuration for clarity.

PREREQUISITES
  • Understanding of electric fields and potentials in capacitors
  • Familiarity with the equation V = E ⋅ s
  • Knowledge of vector directionality in electric fields
  • Ability to interpret physical diagrams related to electric fields
NEXT STEPS
  • Study the concept of electric potential difference in capacitors
  • Learn how to sketch electric field lines and equipotential surfaces
  • Explore the implications of electric field direction on potential calculations
  • Investigate the relationship between electric field strength and potential in different configurations
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone seeking to understand electric potential calculations in capacitor systems.

miyayeah
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Homework Statement


The electric field inside a parallel plate capacitor is measured to be E= -3500 N/C i. The electric potential at point XA = 3.00 m is measured to be 1500V. What is the electric potential at point XB = 0 m?

Homework Equations


V=E⋅s

The Attempt at a Solution


I think I need to relate the given point to the actual point within the electric field, so I did:

s= V/E = (1500V) / (3500N/C) = 2.333 m
I am not how I should go with this problem. Any help would be appreciated!
 
Last edited:
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miyayeah said:
V=E⋅s
is good. So is s= V/E . But V is not -3500 N/C And E is not 1500 V.

Always check the dimensions of what you write down !

Perhaps you can make a sketch showing the positions, E and V ?
 
BvU said:
is good. So is s= V/E . But V is not -3500 N/C And E is not 1500 V.

Always check the dimensions of what you write down !

Perhaps you can make a sketch showing the positions, E and V ?
Thank you for pointing that out. I edited it now.

If I were to sketch, I would draw the electric potential increasing from the negative to the positive end, and the electric field vector going from the positive to the negative end.
I think I am mainly confused about where I should put the positions, because I'm not sure if X=0 mean the position is at the negative plate.
 

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