Charge density - parallel plates

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SUMMARY

The discussion revolves around calculating the electric field intensity and potential difference between two large, parallel conducting plates with a surface charge density of 47.0 nCm-2. The electric field intensity is determined to be 5310.7 N/C using the formula E = σ/ε₀, where σ is the surface charge density and ε₀ is the permittivity of free space. The potential difference between the plates is calculated using the relationship V = Ed, resulting in a value of 116.8 V. The key takeaway is the direct relationship between electric field intensity and potential difference in a uniform field.

PREREQUISITES
  • Understanding of electric field concepts and formulas
  • Familiarity with surface charge density calculations
  • Knowledge of the relationship between electric field and potential difference
  • Basic grasp of electrostatics and conducting plates
NEXT STEPS
  • Study the derivation of the electric field between parallel plates
  • Learn about the concept of surface charge density in electrostatics
  • Explore the relationship between electric field and potential difference in various configurations
  • Investigate applications of electric fields in capacitors and other electrical components
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding electrostatics, particularly the behavior of electric fields between charged conductors.

pat666
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Homework Statement


5. Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm.
a) If the surface charge density for each plate has a magnitude of 47.0 nCm^-2, what is the magnitude of the electric field intensity in the region between the plates?
b) What is the potential difference between the two plates? (2 marks)



Homework Equations





The Attempt at a Solution


The wording here confuses me a bit, is Electric field intensity something different to the electric field?
I've done E=4.7*10^-9/8.85E-12 =530.8N/C but is this the field intensity? Part b I am not sure about, I need some help for it to.

Thanks
 
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Electric field and electric field intensity: same thing. (Check your arithmetic: is the charge density 4.7 or 47?)

For part b: You'll need to use the distance.
 
47, so E=5310.7N/C
Then V=kq/r but I don't know q so how would I do that?
 
pat666 said:
Then V=kq/r but I don't know q so how would I do that?
That formula is for a point charge, which is not relevant here. What's the general relationship between field and potential?
 
found it, V=Ed

so V=116.8V?
 
You got it.
 
Thanks
 

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