# Theory about electric field intensity

1. Dec 16, 2013

### Coco12

1. The problem statement, all variables and given/known data

The electric field intensity at a point between two large parallel plates is 4.8 × 102
N/C. What does the electric
field become, as a result of each of the following changes, considered separately (the plates are isolated),
(a) when the distance between the plates is halved
when the plates are connected together by a conducting wire
2. Relevant equations

E=kq/d^2
E= V/d
E=kq/q

3. The attempt at a solution

There are many different formulas for electric field intensity.
In the first formula, it is inversely proportional to the square of the distances
That would give 4.8* 10^2 divided by 4.
The answer says E is independent of d. Where did they get that?

When the plates are connected , the electric field will be zero because it is connected to one another and would act as a conductor?

2. Dec 16, 2013

### CAF123

If the size of the plates are given to be large relative to the separation between them, then what can be said about the electric field between the plates?

If the plates are connected by a conducting wire, then there cannot be a separation of charge (I.e + charge on one plate and - charge on the other). What does this tell you?

3. Dec 16, 2013

### Coco12

I understand the conducting wire now. However what relation does the size have to do with distance?

4. Dec 16, 2013

### CAF123

If the span of the plates is considerably larger than the distance between them, then the E field is always normal to the plates (we can neglect edge effects which would otherwise 'bend' the electric field). You can then use Gauss' Law to determine this E field. It is a result which you perhaps have in your notes.

5. Dec 16, 2013

### Coco12

We never did gauss law yet

6. Dec 16, 2013

### Coco12

I still don't understand how come electric field intensity is not affected by distance. What about the formula : kq/d^2?

7. Dec 16, 2013

### Coco12

What is the formula for finding electric field intensity between parallel plates?

8. Dec 17, 2013

### hjelmgart

Yeah, I can understand, why you are confused. However, the result can be directly derived from Coulombs law (kq/d^2).

The thing is, we are now dealing with a continuous charge distribution, so when using Coulombs law, you need to sum over every charge. Now we assume, we have an infinite amount of charges, and you get an infinite sum, which after being evaluated gives:

E = σ/2ε

You have most probably seen this equation, perhaps your book doesn't derive it, you should ask your teacher then. It's hard to explain it properly on a forum. Nonetheless, if you aren't much familiar with calculus courses, I would just forget about it, and believe it. You may come across it at a later point. Sometimes it is ok to just accept the general equations, as they are always based on approximations, however, for most uses they prove to be "accurate enough", so we accept it.

However, you may be able to understand it in a simple manner, take this illustration of two point charges. What makes the intensity of the field change, is that the electric field "bends", however, for an infinitely large flat plate of a continuous charge distribution, the field does not bend, given that it is much larger, than the distance between them.