- #1

serverxeon

- 101

- 0

somehow the force isn't just F=qE, where E=Q/2εA

What am I missing here?

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- Thread starter serverxeon
- Start date

In summary, the conversation discusses the task of finding the force exerted by one parallel plate on the other in a capacitor. The speaker mentions that the force is not simply F=qE, where E=Q/2εA, and asks for clarification on what they are missing. The other person suggests using Coulomb's law or calculating the electric field due to a sheet of charge separately. The speaker also mentions that their solution is lacking a coefficient and questions if the electric field is actually 1/2 E. They suggest calculating the field to find the answer.

- #1

serverxeon

- 101

- 0

somehow the force isn't just F=qE, where E=Q/2εA

What am I missing here?

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- #2

Simon Bridge

Science Advisor

Homework Helper

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Erm - direction?What am I missing here?

You can check by using another method ... i.e. Coulombs law for a point charge and add them up, or start with the electric field due to a sheet of charge by itself.

- #3

serverxeon

- 101

- 0

based on my (confused and vague) understanding, the electric field is actually 1/2 E or something?

- #4

Simon Bridge

Science Advisor

Homework Helper

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Is it? Why not *calculate* the field?

- #5

Nickie

- 1,822

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The electrostatic force between two parallel plates in a capacitor is not simply given by the equation F=qE. This equation only applies to a point charge in an electric field, and does not take into account the geometry of the plates. In order to calculate the force between the two plates, you must consider the electric field between them and the area of the plates. This is why the equation for the electric field between parallel plates is E=Q/2εA, where Q is the charge on each plate, ε is the permittivity of the medium between the plates, and A is the area of the plates. To find the force, you would need to multiply this electric field by the charge on one of the plates, and then by the area of the other plate. This will give you the total force exerted by one plate on the other. Additionally, other factors such as the distance between the plates and any dielectric material between them may also affect the force. It is important to consider all relevant factors when calculating the electrostatic force between parallel plates.

Electrostatic force is the attractive or repulsive force between two charged objects. It is caused by the interaction of positive and negative charges, and is governed by Coulomb's Law.

The magnitude of electrostatic force between two charged objects is determined by the product of the charges on the objects and the distance between them. It can be calculated using the equation F = (k * q_{1} * q_{2}) / d^{2}, where k is the Coulomb's constant, q_{1} and q_{2} are the charges on the objects, and d is the distance between them.

As the distance between two charged objects increases, the electrostatic force between them decreases. This relationship follows an inverse square law, meaning that the force decreases by a factor of four when the distance between the objects is doubled.

A dielectric material is an insulating material that can be placed between two charged objects. It reduces the electrostatic force between the objects by decreasing the electric field between them. This is due to the polarization of the dielectric material, which creates an opposing electric field that weakens the overall force.

The electrostatic force on a parallel plate capacitor is directly proportional to the charge on the plates and the distance between them. It can be calculated using the equation F = (q * E) / d, where q is the charge on the plates, E is the electric field between the plates, and d is the distance between the plates.

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