What is the Magnitude of the Electric Field Between Parallel Plates?

In summary, the problem involves two parallel square metal plates with a distance of 1.56 cm between them and sides of length 0.560 m. One plate has a charge of +1.74×10-6 C while the other has a charge of -1.74×10-6 C. The task is to calculate the magnitude of the electric field between the plates, assuming a uniform surface charge distribution. The correct equation to use is E = sigma/epsilon, where sigma is the charge density and epsilon is the permittivity of free space. The value of epsilon is 8.85E-12, and the area of one plate is 0.3136 m^2. Therefore, the total electric
  • #1
badd99
21
0
Problem:

Two square metal plates are placed parallel to each other, separated by a distance d= 1.56 cm. The plates have sides of length L = 0.560 m. One of the plates has charge Q= + 1.74×10-6 C, while the other plate has charge -Q. Calculate the magnitude of the electric field between the plates, not close to the edge, i.e., assume a uniform surface charge distribution

Here's what I have so far: The electric field between them obviously isn't 0.

I've got sigma = (magnitude of charge)/area.
The electric field for one plate is E = sigma/(2 * epsilon).
Since the fields from both plates in between them point in the same direction, the total field would be E = sigma/epsilon. Would I use E=sigma/epsilon OR E=sigma/(2*epsilon) here for the correct answer?

epsilon is just 8.85 here not 8.85E-12 I am 99% but I don't really know why.

How in the WORLD do I find the area to find sigma though?
 
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  • #2
badd99 said:
The electric field for one plate is E = sigma/(2 * epsilon).
Since the fields from both plates in between them point in the same direction, the total field would be E = sigma/epsilon. Would I use E=sigma/epsilon OR E=sigma/(2*epsilon) here for the correct answer?
You want the total field, of course.

epsilon is just 8.85 here not 8.85E-12 I am 99% but I don't really know why.
You're wrong here: It's 8.85E-12.

How in the WORLD do I find the area to find sigma though?
The plates are square. Find the area.
 
  • #3
Im still not getting it right...

I used 1.74×10-6 C as my charge and found the area of the one plate...do I need to find the total area (ie. add the two areas?)

I then got sigma using the above and plugged it into E= sigma / (2* epsilon)

I tried using 8.85 and 8.85E-12 and both incorrect.

Help?
 
  • #4
badd99 said:
I used 1.74×10-6 C as my charge and found the area of the one plate...
Good. What did you get for the area.

do I need to find the total area (ie. add the two areas?)
No.

I then got sigma using the above and plugged it into E= sigma / (2* epsilon)
What did you get for sigma?

I tried using 8.85 and 8.85E-12 and both incorrect.
No point in plugging the wrong value.
 
  • #5
area came out to: .3136 m^2
sigma came out to be: 5.548E-6

Plugged that into E = sigma / (2* epsilon) gave me: 313472.85 N/C and it was incorrect.
 
  • #6
badd99 said:
Plugged that into E = sigma / (2* epsilon) gave me: 313472.85 N/C and it was incorrect.
You want the total field, not just that from one side. Double your answer.
 
  • #7
I thought since we were doing 2*epsilon it was for both plates...aka total field.

I have the answer right now but don't really get why we had to double as I thought it was as stated above.

Thanks for the help!
 
  • #8
badd99 said:
I thought since we were doing 2*epsilon it was for both plates...aka total field.
That factor of 2 is in the denominator, so that version is only half the field. (See my response in post #2.)
 
  • #9
Thank you very much...I posted one more question if you don't mind.

Much appreciated!
 

FAQ: What is the Magnitude of the Electric Field Between Parallel Plates?

What is an electrical field between plates?

An electrical field between plates is a region in space where an electrically charged particle experiences a force due to the presence of electrically charged plates. The strength and direction of the electrical field is determined by the magnitude and distribution of the charges on the plates.

How is the strength of the electrical field between plates calculated?

The strength of the electrical field between plates is calculated by dividing the magnitude of the charge on the plates by the distance between them. This is known as Coulomb's law, which states that the electric field strength is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance between the charges.

What factors affect the strength of the electrical field between plates?

The strength of the electrical field between plates is affected by the magnitude of the charges on the plates, the distance between the plates, and the dielectric material between the plates. A higher magnitude of charge or a smaller distance between the plates will result in a stronger electrical field. The type of dielectric material between the plates can also affect the strength of the field.

What is the direction of the electrical field between plates?

The direction of the electrical field between plates is always from the positively charged plate to the negatively charged plate. This is because like charges repel each other and opposite charges attract each other. Therefore, the positively charged plate repels positive charges and attracts negative charges, creating a net force towards the negatively charged plate.

What are some real-life applications of the electrical field between plates?

The electrical field between plates has many real-life applications, such as in capacitors, which are used in electronic circuits to store and release energy. It is also used in particle accelerators to accelerate charged particles, in photocopiers to create an image on a drum, and in Van de Graaff generators to produce high-voltage electric fields for experiments.

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