Electric field for a capacitor

In summary, the student attempted to solve a problem related to the electric field between two parallel plates using a formula provided in the textbook, but ran into difficulty when accounting for the distance between the plates.
  • #1
vysero
134
0

Homework Statement



I have browsed similar threads and have only found more confusion. My question is if I have two parallel plates with an area A = 1m^2 and they are separated by a distance d = .03m. Each has a charge of Q and -Q respectively where Q = 1nC. I want to find the electric field between the plates.[/B]

Homework Equations



The equation I used was from my textbook: δ/2ε˳[/B]

The Attempt at a Solution



So I said that E(+) = 1nC/2(8.85x10^-12) and that E(tot) = 2(E(+)) = 112.99 n/c. However, this calculation does not take into account the distance between the two plates which I feel it should. Where have I gone wrong here?[/B]
 
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  • #2
You would need to calculate the capacitance - and then use Q = CV.
 
  • #3
You can calculate the field by going the route suggested by Svein, or by using the formula provided. The distance between the plates doesn't matter for the second method provided that the geometry of the capacitor allows the "field is uniform between plates" assumption to hold and what's called "fringe effects" can be neglected. This is almost always the case for problems that are assigned.
 
  • #4
gneill said:
You can calculate the field by going the route suggested by Svein, or by using the formula provided. The distance between the plates doesn't matter for the second method provided that the geometry of the capacitor allows the "field is uniform between plates" assumption to hold and what's called "fringe effects" can be neglected. This is almost always the case for problems that are assigned.

Okay so where did I go wrong when using the provided formula? Or did I do that right?
 
  • #5
Looks right to me. You summed the field contributions of the two plates.
 
  • #6
gneill said:
Looks right to me. You summed the field contributions of the two plates.

Oh :P okay thanks!
 

What is an electric field?

An electric field is a region in space where electrically charged particles experience a force. It is represented by electric field lines that point in the direction of the force.

How is the electric field for a capacitor calculated?

The electric field for a capacitor is calculated by dividing the voltage across the capacitor by the distance between the plates. It is also affected by the dielectric constant of the material between the plates.

Why is the electric field inside a capacitor constant?

The electric field inside a capacitor is constant because the plates of the capacitor are parallel and the distance between them is uniform. This creates a uniform electric field between the plates.

What is the direction of the electric field for a capacitor?

The electric field for a capacitor always points from the positive plate to the negative plate. This is because positive charges are attracted to negative charges and will move in the direction of the electric field lines.

How does the electric field for a capacitor change when the voltage or distance between the plates is altered?

The electric field for a capacitor is directly proportional to the voltage and inversely proportional to the distance between the plates. This means that if the voltage is increased, the electric field will also increase, and if the distance between the plates is decreased, the electric field will also increase. The opposite is also true.

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