# Energy Density with a Dielectric

• racecar_
In summary, the conversation discusses the confusion about how the electric field changes in the given problem and whether the equation E' = E/Ke=E/2 is applicable. It is determined that this equation is not valid in this situation and the correct equation is E = V/d. The topic of energy density is also brought up and it is mentioned that the equation (1/2)εE^2 is not applicable here. Instead, the approach of finding the stored energy and dividing by the volume is suggested as a better method. It is also mentioned that using ε for both vacuum and dielectric is a key mistake. The conversation concludes with the confirmation that the electric field will remain the same and a question about what changes in the equation for
racecar_
Homework Statement
Two parallel plates, having a vacuum in between them, are separated by d=0.01 m apart and are connected to a battery maintaining 9V between the plates and a charge of Q0 magnitude on each plate. The separation between the plates is then raised to 2d, the battery voltage is increased to 18V, and a dielectric is added with dielectric constant 2. What is the energy density difference after the addition of the dielectric?
Relevant Equations
energy density = (1/2)εE^2
I am confused about how the electric field changes in this problem - is E' = E/Ke=E/2? Is E = V/d a correct usage?

When I solve it this way, the answer is incorrect:
change in energy density = (1/2)ε(E'2- E2) = (1/2)ε(E2/4 - E2) = (1/2)ε(-3/4)(V/2d)2.

What am I doing wrong? Thanks.

racecar_ said:
I am confused about how the electric field changes in this problem - is E' = E/Ke=E/2?
This equation does not apply in your situation. This equation would be valid if, after charging the capacitor, you disconnect the capacitor from the battery and then add the dielectric.

racecar_ said:
Is E = V/d a correct usage?
Yes. This is generally valid for any parallel plate capacitor.

racecar_
The obvious approach is to calculate the stored energy in each case and divide by the volume.

racecar_, nasu and TSny
TSny said:
This equation does not apply in your situation. This equation would be valid if, after charging the capacitor, you disconnect the capacitor from the battery and then add the dielectric.
So the electric field will stay the same? For the equation energy density = (1/2)εE^2, what changes then?

kuruman said:
The obvious approach is to calculate the stored energy in each case and divide by the volume.
Yeah, I think it's a better approach. In what situations should I use finding the stored energy and dividing the volume rather than (1/2)εE^2, and vice versa? Thanks.

Either approach should work if you know what you’re doing. Here you made a key mistake by using ε for both vacuum and dielectric between the plates instead of for only the dielectric.

racecar_ said:
So the electric field will stay the same?
Yes.

racecar_ said:
For the equation energy density = (1/2)εE^2, what changes then?
Following up a little on @kuruman's post. The energy density of an electric field in a vacuum is ## \frac1 2 \varepsilon_0 E^2##. This expression is modified for an electric field inside a dielectric. Have you covered this in your course?

If not, then use the approach of finding the stored energy divided by the volume.

## What is energy density with a dielectric?

Energy density with a dielectric is a measure of the amount of energy stored in an electric field per unit volume of a material. It takes into account the presence of a dielectric material, which can increase or decrease the energy density compared to a vacuum.

## How is energy density with a dielectric calculated?

The energy density with a dielectric is calculated using the equation U = 1/2 * ε * E^2, where U is the energy density, ε is the permittivity of the material, and E is the electric field strength.

## What is the role of a dielectric in energy density?

A dielectric material is an insulating material that can store electric charge. In the context of energy density, a dielectric can increase or decrease the energy density depending on its permittivity. A higher permittivity material will have a higher energy density, while a lower permittivity material will have a lower energy density.

## How does energy density with a dielectric affect electric fields?

Energy density with a dielectric has a direct relationship with electric fields. As the energy density increases, the electric field strength also increases. This is because the energy stored in the electric field is proportional to the square of the electric field strength.

## What are some real-world applications of energy density with a dielectric?

Energy density with a dielectric is important in various technologies such as capacitors, batteries, and photovoltaic cells. It is also crucial in understanding the behavior of materials in electric fields, which is relevant in fields such as electronics, telecommunications, and energy storage.

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