Electric Field Problem: Solving for E in a System of Oppositely Charged Plates

[Prashasti]

Question : Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0*10^-22. What is E (Electric field) : (a) in the outer region of the first plate, (b) in the outer region of the second plate?

Answer: According to the book, answers are (a) 0 (b) 0 And the reason they've given is - "Net electric field is Zero because electric fields due to both the plates will cancel out each other".

My objection : How? How do they cancel each other's effect when that particular region is NOT EQUIDISTANT from each other, and as Electric field varies inversely with the square of distance, HOW ARE THEY EQUAL IN MAGNITUDE?

 

[Simon Bridge]

...as Electric field varies inversely with the square of distance

... this is only correct for sherical charge distributions - i.e. a sphere or a point charge.

Why is it important that the plates are "large"?

The electric field from two sheets of charge like that is the sum of the fields due to the single sheets alone.
What is the electric field due to a single large sheet of charge?

 

[Prashasti]

I got it. Thank you so much. I think it is E = sigma/2×vacuum permittivity?!

 

[Simon Bridge]

Well done - yeah: the field due to a large plate (except near the edges) is uniform.
Since the plates have opposite signs, their fields cancel everywhere except in between them (and by the edges) where they add together.

 

[sardar]

As a scientist, it is important to consider all factors when solving a problem. In this scenario, we must take into account the distance between the plates and the surface charge densities on each plate. The electric field is a vector quantity, meaning it has both magnitude and direction. In this case, the direction of the electric field will be from the positively charged plate to the negatively charged plate.

To solve for the electric field in the outer region of the first plate, we can use the formula E = σ/ε0, where σ is the surface charge density and ε0 is the permittivity of free space. Plugging in the given values, we get E = (17.0*10^-22 C/m^2) / (8.85*10^-12 C^2/N*m^2) = 1.92*10^-10 N/C. This tells us the magnitude of the electric field at any point in the outer region of the first plate.

Similarly, we can solve for the electric field in the outer region of the second plate using the same formula. However, since the surface charge density on the second plate is of opposite sign, the direction of the electric field will be opposite to that of the first plate. This means that the magnitude of the electric field will be the same, but the direction will be reversed.

Now, to address the concern about the distance between the plates, it is true that the electric field varies inversely with the square of distance. However, in this scenario, we are looking at the outer region of each plate, which is at a considerable distance from the plates themselves. This distance is large enough that the electric field from each plate can be considered to be constant and uniform in that region. Therefore, the electric fields from each plate can be added together, resulting in a net electric field of 0. This is because the electric fields are equal in magnitude but opposite in direction, thus canceling each other out.

In conclusion, the electric field in the outer region of both plates is 0 because the electric fields from each plate cancel each other out due to their equal magnitude and opposite direction. This is possible because the distance between the plates is large enough that the electric fields can be considered constant and uniform. It is important to carefully consider all factors and use the appropriate formulas when solving problems in order to arrive at accurate solutions.