Changes to electric field between two parallel plates

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SUMMARY

The electric field intensity between two parallel plates is initially 4.8 x 10^2 N/C. When the charge on each plate is tripled, the electric field intensity also triples, resulting in an electric field of 1.44 x 10^3 N/C. In the case where the plates are connected by a conducting wire, the electric field becomes zero due to the equalization of potential across the plates, eliminating the potential difference.

PREREQUISITES
  • Understanding of electric field concepts
  • Familiarity with Coulomb's Law
  • Knowledge of potential difference and its effects
  • Basic principles of electrostatics
NEXT STEPS
  • Study the relationship between charge and electric field using the equation E = kQ/d²
  • Explore the effects of connecting charged plates with a conducting wire
  • Learn about the concept of equipotential surfaces in electrostatics
  • Investigate the principles of electric field superposition
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Students studying physics, particularly those focusing on electromagnetism, as well as educators seeking to clarify concepts related to electric fields and charge interactions.

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Homework Statement



The electric field intensity at a point between two large parallel plates is 4.8 x 10^2 N/C.
What does the electric field become as a result of each of the following changes, considered separately:

a) when the amount of charge on each plate is tripled
b) when the plates are connected together by a conducting wire

for a, i think the electric field intensity would be tripled as well? i don't know which equation to look at to find the answer -- ε = kQ/r^2 doesn't seem right because there are two charges, one on each plate, and ε = V/d doesn't work because d doesn't come into play. i don't even have a guess for b). any explanation anyone could provide would be appreciated.
 
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a)How can you obtain the electric field given a highly symmetrical charge configuration?
b)What do you expect to happen when there is potential difference across a conducting wire?
 

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