What Determines Electric Field Strengths of Two Charged Rods

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SUMMARY

The discussion focuses on calculating the electric field strengths generated by a 10cm long thin glass rod uniformly charged to 10nC and a 10cm long thin plastic rod uniformly charged to -10nC, positioned 4cm apart. The key takeaway is that the problem requires the use of the electric field formula for a finite line of charge rather than treating the rods as capacitors, as neither material is a conductor. The relevant equation for electric field strength is E = η/ε0 = (4πK⋅Q) / A, but the area is not applicable in this scenario.

PREREQUISITES
  • Understanding of electric fields and charge distributions
  • Familiarity with the concept of finite line of charge
  • Knowledge of electrostatics, specifically Coulomb's law
  • Basic calculus for integrating electric field contributions
NEXT STEPS
  • Study the electric field due to a finite line of charge using the formula provided in the linked document
  • Learn about superposition of electric fields from multiple charge distributions
  • Explore the differences between conductors and insulators in electrostatics
  • Review the principles of capacitors and their requirements for charge storage
USEFUL FOR

Students studying electrostatics, physics educators, and anyone interested in understanding electric field calculations involving charged objects.

miyayeah
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Homework Statement


The problem I encountered goes like this: A 10cm long thin glass rod uniformly charged to 10nC and a 10cm long thin plastic rod uniformly charged to -10nC are placed side by side 4cm apart. What are the electric field strengths E1 to E3 at distances 1cm, 2cm, and 3cm from the glass rod along the line connecting the midpoints of the two rods?

Homework Equations


E= η/ε0 = (4πK⋅Q) / A

The Attempt at a Solution


I am not quite confused about getting the answer to this problem but as to how I should approach the problem. I thought that the question was describing a capacitor, since the two rods are charged equally but oppositely. But when I tried to use the formula (as shown above), I realized I have no way to find the area. So, I realized that it was not a capacitor. How would I know, aside from looking at the formula, that the question is not describing a capacitor? Or when it describes a capacitor?
 
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To have a capacitor, you need two conductors near one another carrying equal and opposite charges. That's not what you have here because neither glass nor plastic can be considered as conductors. You need to find the electric field due to finite line of charge and superimpose the two fields.
 
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kuruman said:
To have a capacitor, you need two conductors near one another carrying equal and opposite charges. That's not what you have here because neither glass nor plastic can be considered as conductors. You need to find the electric field due to finite line of charge and superimpose the two fields.
Thank you for your explanation. Which formula should I use if I am solving for the electric field due to finite line of charge?
 

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