5. Energy of a Capacitor in the Presence of a Dielectric

In summary, a capacitor is an electrical component that stores energy in an electric field. It consists of two conductive plates separated by an insulating material, known as a dielectric. The energy stored in a capacitor is equal to half of the product of its capacitance and the square of the voltage across it. A dielectric material is an insulating material that is placed between the plates of a capacitor, helping to increase its capacitance and energy storage capacity. The presence of a dielectric material also affects the energy stored in a capacitor, as it allows for a higher voltage to be applied without causing breakdown. The energy of a capacitor is directly proportional to the dielectric constant of the material between its plates, with a higher dielectric constant
  • #1
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Homework Statement


A dielectric-filled parallel-plate capacitor has plate area of 25.0 cm^2 and plate separation of 9.00 mm. The capacitor is connected to a battery that creates a constant voltage of 7.50 V. The dielectric constant is 2.00. Throughout the problem, use 8.85×10−12 C^2/N \cdot m^2


Homework Equations


U=1/2C*V^2
C=[tex]\epsilon[/tex]A/d
C=K*C_0



The Attempt at a Solution



2.07×10^−10
Keep in mind that the area of each portion of the capacitor is half that of the original capacitor.
 
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  • #2
C=\epsilonA/dC=(2)(8.85 × 10^−12) (25.0 \ cm^2)/(0.009 \ m)=2.07×10^−10 U=1/2C*V^2U=1/2(2.07×10^−10 )(7.5^2) = 8.59×10^−6 J
 
  • #3


I would first calculate the capacitance of the capacitor in its original state, without the dielectric. Using the equation C = εA/d, where ε is the permittivity of free space (8.85×10−12 C^2/N⋅m^2), A is the plate area (25.0 cm^2), and d is the plate separation (9.00 mm), I would find that the capacitance is 2.78×10−11 F.

Next, I would use the equation C = K*C_0 to find the new capacitance with the dielectric, where K is the dielectric constant (2.00) and C_0 is the original capacitance. This would give a new capacitance of 5.56×10−11 F.

Finally, I would use the equation U=1/2C*V^2 to calculate the energy of the capacitor in the presence of the dielectric. Plugging in the new capacitance and the constant voltage of 7.50 V, I would find that the energy of the capacitor is 2.07×10^−10 J.

I would also note that the area of each portion of the capacitor is half of the original area, since the dielectric fills the space between the plates. This information may be useful for future calculations or experiments involving this capacitor.
 

Related to 5. Energy of a Capacitor in the Presence of a Dielectric

1. What is a capacitor?

A capacitor is an electrical component that stores energy in an electric field. It consists of two conductive plates separated by an insulating material, known as a dielectric.

2. What is the energy stored in a capacitor?

The energy stored in a capacitor is equal to half of the product of its capacitance and the square of the voltage across it. It is given by the formula E = 1/2 * C * V^2, where C is the capacitance and V is the voltage.

3. What is a dielectric material?

A dielectric material is an insulating material that is placed between the plates of a capacitor. It helps to increase the capacitance of the capacitor by reducing the electric field between the plates.

4. How does the presence of a dielectric affect the energy of a capacitor?

The presence of a dielectric material between the plates of a capacitor increases the energy stored in the capacitor. This is because the dielectric material reduces the electric field between the plates, allowing for a higher voltage to be applied to the capacitor without causing breakdown.

5. What is the relationship between the energy of a capacitor and the dielectric constant?

The energy stored in a capacitor is directly proportional to the dielectric constant of the material between its plates. A higher dielectric constant means a higher energy storage capacity for the capacitor.

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