Parallel Conducting plates problem

Click For Summary
When one of the two parallel conducting plates is charged with +Q while the other remains neutral, the charged plate will induce a negative charge on the face of the neutral plate closest to it. The outer face of the charged plate will have a charge of +Q, while the outer face of the neutral plate will remain neutral, resulting in a total charge distribution of +Q on the inner face of the charged plate and -Q on the inner face of the neutral plate. The electric field at point A (outer side of plate 1) will be zero, at point B (between the plates) will be uniform and directed from the positively charged plate to the negatively charged plate, and at point C (outer side of plate 2) will also be zero. The equations governing the charge distribution can be derived from the total charge conservation and the condition that there is no electric field inside the conductors. Understanding these principles clarifies the behavior of electric fields and charges in parallel plate configurations.
ajmaeen
Messages
1
Reaction score
0
If we have two parallel conducting plates ( of surfaces 1msq each), set at a very small distance apart, so that the plates can be considered like infinite plates (no edge effect).

Both plates are neutral at first.

Now a charge (say +Q) is placed on one of the plates. the other still is neutral

What will be the charge on each face of both the plates, and the field at A, B and C, where A is outer side of plate 1, B is middle of plate 1 and 2, and C in outer side of plate 2?


My thoughts:
i know that if the other plate also had an opposite excess charge then on the outer sides charges are 0 , while inside they would be +Q and -Q. so field would also only exist in the middle.

but I am a but confused on what it should be incase of only one plate with the excess charge?
 
Physics news on Phys.org
Create unknowns for the four surface charges. You can immediately write down two equations based on knowing the total charge in each plate. Since there is no field inside either plate, you can develop two more equations.
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

  • · Replies 31 ·
2
Replies
31
Views
1K
Replies
11
Views
3K
  • · Replies 18 ·
Replies
18
Views
3K
Replies
1
Views
1K
Replies
6
Views
1K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 58 ·
2
Replies
58
Views
5K
  • · Replies 20 ·
Replies
20
Views
2K