# Problem in orienting two plates as instructed in question

1. Sep 15, 2015

### gracy

1.The problem statement, all variables and given/known data
Two large thin metal plates are parallel and close to each other. on their inner faces ,the plates have surface charge densities of opposites signs and magnitude 1.77 multiplied by 10^-11 coulomb per square meter.What is electric field
a)to the left of the plates
b)to the right of the plates
c)in between the plates
2. Relevant equations
E=σ/2ε0

3. The attempt at a solution
What is the proper orientation when the question says Two large thin metal plates are parallel and close to each other
I thought the following is correct

But this is not correct.
The answer is Fields due to both the plates will be equal and opposite so net field will be zero.I can not differentiate between the cases that are
to the left of the plate (2) and between the plates

2. Sep 15, 2015

### TSny

In part (a), "to the left of the plates" refers to the far left region; that is, the region to the left of the left plate.
Likewise in part (b), "to the right of the plates" refers to the far right region; that is, the region to the right of the right plate.

3. Sep 15, 2015

### gracy

Even if that's the case how come

4. Sep 15, 2015

### gracy

I mean
Can electric field due to plate (2)go beyond the plate (1) i.e to the far left region
or even if it can ;would the electric field intensity of the charges of plate 2 be equal to that of plate one there in order to cancel the effect?

5. Sep 15, 2015

### ehild

Imagine one plate: When it is positively charged, the field lines emerge from it, and point away from the plate at both sides.
If the charge is negative, the field lines enter into the plate at both sides.
You can substitute the plates by planar distributions of charges. Both planes have their field everywhere, as shown in the picture. The orange lines are due to the positive charges, the blue lines belong to the negative ones. Apply he Superposition Principle: the resultant field is the sum of the fields of the individual planes.

6. Sep 15, 2015

### gracy

7. Sep 15, 2015

### ehild

If the charges per unit surface area are equal, the fields are of equal magnitude.

8. Sep 15, 2015

### gracy

But electric field also depends on r i.e distance from source charge

9. Sep 15, 2015

### ehild

NO, if the plates are infinite (at least their lateral sizes are much larger than the distance between them). You wrote that E=σ/(2ε0) !

10. Sep 15, 2015

### gracy

YES .I wrote that but I did not understand.

11. Sep 15, 2015

### gracy

But it's beyond them not between them

12. Sep 15, 2015

### gracy

13. Sep 15, 2015

### ehild

It is symmetry again. The charge is evenly distributed on the plane. At a point the fields of the individual charges add as vectors. The parallel components cancel and only the perpendicular components remain. As the plane is of infinite extension, the field is the same above each points. If you integrate the contribution of all charges you will get the same result, independently how far the point is from the plate.

14. Sep 15, 2015

### ehild

The field of an infinite plate with even surface charge distribution is the same at both sides, only of opposite sign.

15. Sep 15, 2015

### gracy

Ok.One more question If these are metal plates ,why charges are spread throughout the surface;charges should reside on borders/outer surface

16. Sep 15, 2015

### ehild

The plates were assumed of infinite sizes.
Even in case of real metal plates, not all charge reside along the borders. But it is true that they are spread on the surfaces.

17. Sep 16, 2015

### gracy

18. Sep 16, 2015

### gracy

19. Sep 16, 2015