# Dielectric inserted in parallel plate capacitor

• Jahnavi
In summary, the force between the plates is given by (1/2)QE .When dielectric is inserted C changes to "kC" .V doesn't change as the plates are connected to battery .
Jahnavi

## The Attempt at a Solution

The force between the plates is given by (1/2)QE ,where Q is the charge on the plates and E is the electric field between the plates .

Now Q = CV and E = V/d , where C is the capacitance , V is the potential difference and d is the plate separation . So , force can be written as F = (1/2)(CV)(V/d) .When dielectric is inserted C changes to "kC" .V doesn't change as the plates are connected to battery .

So , force F increases by a factor "k'" to become kF on insertion of the dielectric while plates are connected to battery .

Is my work correct ?

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Jahnavi said:
The force between the plates is given by (1/2)QE ,where Q is the charge on the plates and E is the electric field between the plates .
I think you need to be careful with calling E the electric field between the plates when there is a dielectric between the plates. The force on Q of one of the plates depends on the value of the field at the surface of the plate where Q is located. Is this value of E the same as the value of the field inside the dielectric? For instance, consider the case where there is a thin layer of vacuum between each plate and the dielectric material. In this case, the field at Q would be the field in the vacuum layer between the plate and the dielectric. This would be true no matter how thin you make the vacuum layer.

Jahnavi
So , if E is the electric field at the surface of plates and E' is electric field between the plates , then E' = E/k and V = E'd .This gives E = Vk/d .

Force F = (1/2)(CV)(E) .After inserting the dielectric , C changes to kC and E could be written as Vk/d . So , basically force increases by a factor k2 .

Is that correct ?

Jahnavi said:
So , if E is the electric field at the surface of plates and E' is electric field between the plates , then E' = E/k and V = E'd .This gives E = Vk/d .

Force F = (1/2)(CV)(E) .After inserting the dielectric , C changes to kC and E could be written as Vk/d . So , basically force increases by a factor k2 .

Is that correct ?
I think that's correct.

Jahnavi
I think it is not wise to use the expression F = (1/2)QE .Instead F = Q2/2Aε0 looks easier to analyse . Using Q= CV , when battery is connected and dielectric is inserted , C increases by a factor k and so does the charge Q . Clearly F increases by a factor k2 .

In case when battery is not connected Q remains same and so force remains unchanged .

Easier to analyse. Isn't it ?

Jahnavi said:
I think it is not wise to use the expression F = (1/2)QE .Instead F = Q2/2Aε0 looks easier to analyse . Using Q= CV , when battery is connected and dielectric is inserted , C increases by a factor k and so does the charge Q . Clearly F increases by a factor k2 .

In case when battery is not connected Q remains same and so force remains unchanged .

Easier to analyse. Isn't it ?
Yes. That's a nicer way to look at it.

Jahnavi
Thanks !

## 1. What is a dielectric material?

A dielectric material is a non-conductive substance that is placed between the plates of a parallel plate capacitor to increase its capacitance. It is typically made of an insulating material such as glass, plastic, or ceramic.

## 2. How does a dielectric material affect the capacitance of a parallel plate capacitor?

A dielectric material increases the capacitance of a parallel plate capacitor by reducing the electric field between the plates. This is due to the polarization of the dielectric material, which creates an opposing electric field and allows for more charge to be stored on the plates.

## 3. What is the dielectric constant of a material?

The dielectric constant, also known as the relative permittivity, is a measure of a material's ability to store electrical energy in an electric field. It is the ratio of the capacitance of a capacitor with the dielectric material to the capacitance of the same capacitor without the dielectric material.

## 4. How does the dielectric constant affect the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor is directly proportional to the dielectric constant of the material inserted between the plates. A higher dielectric constant means a higher capacitance and vice versa.

## 5. Can any dielectric material be used in a parallel plate capacitor?

No, not all dielectric materials are suitable for use in a parallel plate capacitor. The material must have a high dielectric constant and low conductivity to effectively increase the capacitance. Additionally, the material must be able to withstand the electric field and not breakdown under high voltage.

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