# Electrical Force between plates(capacitor)

• haleycomet2
In summary, the conversation discusses the equations for measuring electrical field strength, E, which can be counted using either E=V/d or E=Qq/4πεr^2. It is noted that in a circuit with a capacitor, the potential difference, V, is determined by the battery and therefore the first equation is not affected by permittivity (dielectric material), while the second equation is affected. The group then discusses the correct explanation for the electrical force between plates, including the formula E=V/d and its derivation. It is also mentioned that the charge Q is for each plate and not the total charge on both plates. The topic of measuring E inside a material is also brought up, with the conclusion that it

#### haleycomet2

we know that electrical field strength,E,can be counted by equation
1. E=V/d,or
2. E=Qq/4$$\pi$$$$\epsilon$$r 2
In the circuit with capacitor, the potential difference ,V is decided by battery.So from the first equation ,the E is not affected by permittivity(dielectric material) but it does affected according to second equation.
What is the correct explanation of the electrical force between plates??

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Welcome to PF. Quite an interesting question! Apparently inserting a dielectric between the plates decreases the E field between the plates, so E = V/d does not apply . . . or you can say that an opposing E field due to the separation of charges in the dielectric is subtracted from the V/d one. See http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dielec.html#c3.

There is something wrong with your second formula - it has the dimensions of a force rather than an electric field.

This is an Electric Field around a point charge: E = k(q/r²)

Go back and check where you got this formula: E = V/d and how it is derived.

* Basically the epsilon is in the Electric Potential (V). V for point charge is V = - integral(E dl)

I went to check the derivation of the formula E=V/d.Referring to the graph,the work done to move charge A to B can be expressed by:
W=qEd or W=qVAB
Therefore by eliminating q,VAB=Ed,so E=VAB/d

Besides,i also think other derivation,in the circuit consist of capacitor,using the formulas below:
E=$$\sigma$$ /$$\epsilon$$
C=$$\epsilon$$ A/d
Q=CV
$$\sigma$$ =Q/A

E=$$\sigma$$ /$$\epsilon$$
=(Q/A)/$$\epsilon$$
=(CV/A)/$$\epsilon$$
=($$\epsilon$$ AV/Ad)/$$\epsilon$$
=V/d

Derivation above showed that the electric field strength is independent of $$\epsilon$$,dielectric material!?If according to the field strength formula E=kq/r2,it seems that the field strength is independent of capacitor area?!
ps:The charge Q counted from Q=CV is the amount of charge on one side of plate or both side of the capacitor?

Thank you again...=)

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E=kq/r² does not apply directly to parallel plates. It is for point or spherical charges.

Q is the charge on each plate.

Your argument for E = V/d is quite convincing. Wikipedia
http://en.wikipedia.org/wiki/Dielectric
says a dielectric reduces the field and increases the charge, but in another article
http://en.wikipedia.org/wiki/Capacitor
they use E = ρ/ε which suggests to me that increasing charge density and increasing ε would leave E unchanged.

Anyway, how would you measure E inside a material?

Oh,thanks ,this is what i think now:
The presence of the dielectric will cause the electric field become weaker,so more charges are transferred to plates until reach the initial electric field strength(same with battery's voltage)again.Therefore,the charge density on the plate is increased,but the electric field(E=V/d)is still the same.

Is it??

can I measure the electric field strength by passing a wire between the plates and measure the force exert on the wire(the flowing electron actually) ,or the effect is too tiny to be observed?

The wire trick won't work. There is no net charge on a wire with current flowing. Anyway, how would you get the wire through the dielectric?

Your latest theory about the E field may well be right. I don't know.

## 1. What is electrical force between plates?

The electrical force between plates, also known as the capacitor force, is the force that exists between two parallel plates when they have opposite charges. It is caused by the attraction between the positive and negative charges on the plates.

## 2. How is the electrical force between plates calculated?

The electrical force between plates is calculated using the formula F = Q1Q2/d^2, where Q1 and Q2 are the charges on the plates and d is the distance between the plates. This formula is known as Coulomb's law.

## 3. What factors affect the electrical force between plates?

The electrical force between plates is affected by the distance between the plates, the amount of charge on each plate, and the dielectric constant of the material between the plates. Increasing the distance or decreasing the charge or dielectric constant will decrease the electrical force.

## 4. What is the unit of measurement for electrical force between plates?

The unit of measurement for electrical force between plates is Newtons (N), which is the same unit used for other types of force. This unit is named after Sir Isaac Newton, a famous scientist who made significant contributions to the field of physics.

## 5. How is the electrical force between plates used in technology?

The electrical force between plates is used in technology to store and release electrical energy. Capacitors, which consist of two plates separated by a dielectric material, are commonly used in electronic circuits to store and regulate electrical charge. They are also used in flash photography, power factor correction, and many other applications.