Diagram of electric potential difference in plate capacitor

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SUMMARY

The discussion centers on calculating the electric potential difference in a plate capacitor filled with two dielectrics with constants 3 and 4, and a metal component. The electric field in the first dielectric is specified as 30 kV/cm. Participants analyzed the implications of this electric field on the voltage across the capacitor and the relationship between the two capacitances. The reference point for potential was established at point C, leading to a necessary adjustment of the potential graph by 120 kV.

PREREQUISITES
  • Understanding of electric fields and potential difference
  • Knowledge of dielectric constants and their effects on capacitance
  • Familiarity with capacitor circuit diagrams
  • Ability to interpret and manipulate voltage graphs
NEXT STEPS
  • Study the relationship between electric field strength and voltage in capacitors
  • Learn about the effects of different dielectric materials on capacitance
  • Explore the concept of electric potential and reference points in circuit analysis
  • Investigate the behavior of electric fields in multi-layer dielectric systems
USEFUL FOR

Students studying electromagnetism, electrical engineers, and anyone involved in capacitor design and analysis.

PhanicKnight
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Homework Statement


Plate capacitor is filled with two dielectrics of constants 3 and 4, and with a metal as seen in the picture. If the el. field in the first dielectric is 30 kV/cm, do:

a) diagram of el. potential difference with calculated characteristic values using point C as a reference point.

b) voltage UBF

2. The attempt at a solution
scan0008.jpg


There's no way φd can be 0 and I don't know any other way of getting to it.
 
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PhanicKnight said:
If the el. field in the first dielectric is 30 kV/cm,
What does this tell you about the voltage across the first capacitor?
PhanicKnight said:
two dielectrics of constants 3 and 4,
What does this tell you about the ratio of two capacitances?
Edit: I just saw your attempt below the question. It wasn't clearly visible on my phone earlier.
 
PhanicKnight said:
There's no way φd can be 0
Why? What can you say about the E-field between C and D?
 
cnh1995 said:
Why? What can you say about the E-field between C and D?
Okay, so φd actually is 0. I tried something (see picture underneath), but I'm not sure if it's correct as I don't have correct answers.

scan0009.jpg
 
The question says to use the point C as the reference point, so presumably you must set that as the zero reference for potential. That means at the very least you need to shift your graph downwards by 120 kV.

How do you explain the dip in potential from D to E? Does the field change directions at E? I don't see a charge there that would account for it.
 
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I solved it, thank you all for the help!
upload_2016-11-16_1-38-17.png
 
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