Changes to electric field between two parallel plates

AI Thread 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 is expected to also triple, but clarification on the correct equations is needed. The equations ε = kQ/r^2 and ε = V/d are not applicable in this scenario due to the configuration of the plates. Connecting the plates with a conducting wire will eliminate the potential difference, resulting in the electric field dropping to zero. Understanding the behavior of electric fields in symmetrical charge configurations and the effects of connecting conductors is crucial for solving these problems.
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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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