# A dielectric-filled parallel-plate capacitor has plate area

• onestar_
In summary: Now, use your expression for U2 and the value of C3 to compute U3, the energy when the dielectric is completely removed.You should find that U3 = U2 + W, where W is the work done by the external agent. If this is not the case, please show your work for computing W.In summary, the problem involves finding the new energy, U3, of a dielectric-filled parallel-plate capacitor after the dielectric has been completely removed while disconnected from a battery. The capacitor has a plate area of 15.0 cm^2, plate separation of 9.00 mm, and dielectric constant of 4.00. The battery creates a constant voltage of 15.
onestar_

## Homework Statement

A dielectric-filled parallel-plate capacitor has plate area A = 15.0 cm2 , plate separation d = 9.00 mm and dielectric constant k = 4.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V . Throughout the problem, use ϵ0 = 8.85×10^−12 C2/N⋅m2 .

U1 = 6.64×10^−10 J
U2 = 4.15×10−10J

A)The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3.

B)In the process of removing the remaining portion of the dielectric from the disconnected capacitor, how much work W is done by the external agent acting on the dielectric?

## The Attempt at a Solution

it says that both my w and U_3 are incorrect so i tried it again by doing it differently such as:

Q=C2*V=3.16*10^-12*15 =4.74*10^-11 C

C3 =eoA/d =(8.85*10^-12)(15*10^-4)/(9*10^-3)

C3=1.475*10^-12 F

U3 =(1/2)(Q^2/C3) =(1/2)(4.74*10^-11)^2/(1.475*10^-12)

U3=7.616*10^-10 J

(but it still continues to tell me i am incorrect) so what am i doing wrong? what is the correct answer?
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and same goes for W i re did it and did:

W=U3-U2 =7.616-10^-10-4.15*10^-10

W=3.466*10^-10 J

Hello, and welcome to PF!

onestar_ said:
... i tried it again by doing it differently such as:

Q=C2*V=3.16*10^-12*15 =4.74*10^-11 C

C3 =eoA/d =(8.85*10^-12)(15*10^-4)/(9*10^-3)

C3=1.475*10^-12 F

Shouldn't C2 be the total capacitance, Ceq, when the dielectric is half-way out? The value you have here does not look correct.

Your value of C3 looks good.

## 1. What is a dielectric-filled parallel-plate capacitor?

A dielectric-filled parallel-plate capacitor is a type of capacitor that consists of two parallel conductive plates separated by a material called a dielectric. This dielectric material is an insulator that can store electrical energy when placed between the plates of a capacitor.

## 2. How does the plate area affect the capacitance of a dielectric-filled parallel-plate capacitor?

The plate area directly affects the capacitance of a dielectric-filled parallel-plate capacitor. The larger the plate area, the higher the capacitance will be. This is because a larger plate area allows for more charges to be stored between the plates, resulting in a higher capacitance.

## 3. What is the formula for calculating the capacitance of a dielectric-filled parallel-plate capacitor?

The formula for calculating the capacitance of a dielectric-filled parallel-plate capacitor is C = εA/d, where C is the capacitance, ε is the permittivity of the dielectric material, A is the plate area, and d is the distance between the plates.

## 4. What is the purpose of using a dielectric material in a parallel-plate capacitor?

The purpose of using a dielectric material in a parallel-plate capacitor is to increase the capacitance of the capacitor. The dielectric material reduces the electric field between the plates, allowing for more charges to be stored on the plates and increasing the capacitor's capacitance.

## 5. How does the dielectric constant of the material affect the capacitance of a dielectric-filled parallel-plate capacitor?

The dielectric constant, also known as the relative permittivity, is a measure of how well a material can store electrical energy. The higher the dielectric constant, the higher the capacitance of the capacitor will be. This is because a higher dielectric constant allows for more charges to be stored on the plates, resulting in a higher capacitance.

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