Changes to electric field between two parallel plates

  • Thread starter chartreuse
  • Start date
  • #1

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.
 

Answers and Replies

  • #2
Defennder
Homework Helper
2,591
5
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?
 

Related Threads on Changes to electric field between two parallel plates

Replies
5
Views
52K
Replies
6
Views
1K
Replies
4
Views
4K
Replies
7
Views
5K
Replies
8
Views
2K
Replies
1
Views
562
Replies
1
Views
1K
Replies
1
Views
3K
Replies
4
Views
34K
Top