Question

Part A

Find the energy U0 stored in the capacitor. Express your answer in terms of A, d, V, and ϵ0. Remember to enter ϵ0 as epsilon0.

Part B

The capacitor is now disconnected from the battery, and the plates of the capacitor are then slowly pulled apart until the separation reaches 3d. Find the new energy U1 of the capacitor after this process.

Express your answer in terms of A, d, V, and ϵ0.

Part C

The capacitor is now reconnected to the battery, and the plate separation is restored to d. A dielectric plate is slowly moved into the capacitor until the entire space between the plates is filled. Find the energy U2 of the dielectric-filled capacitor. The capacitor remains connected to the battery. The dielectric constant is K.

Express your answer in terms of A, d, V, K, and ϵ0.

Answer 1

Concepts and reason

The main concepts required to solve this problem are the potential energy, capacitance, potential difference, area, and distance.

Initially, write the equation for the energy stored in the capacitor and the equation for the capacitance of the parallel plate capacitors. Use these equations and find the energy stored in the capacitor for the different cases given in the question.

Fundamentals

The equation for the energy stored in the capacitor is, \(U=\frac{1}{2} C V^{2}\)

Here, \(C\) is the capacitance of the capacitor, and \(V\) is the potential difference between the parallel plate capacitors. The equation for the capacitance in the parallel plate capacitors is, \(C=\frac{\varepsilon_{o} A}{d}\)

Here, \(\varepsilon_{o}\) is the permittivity of free space, \(A\) is the area of the capacitor plates, and \(d\) is the separation of the capacitor plates.

(A) Let the separation of the plates of the capacitors be \(\mathrm{d}\). The equation for the capacitance of the parallel plate capacitor is,

\(C=\frac{\varepsilon_{o} A}{d} \ldots \ldots(1)\)

Here, \(\mathrm{d}\) is the separation of the two plates. The equation for the energy store in the capacitors is,

\(U_{o}=\frac{1}{2} C V^{2}\)

Substitute the equation (2) in above equation (1).

\(U_{o}=\frac{1}{2}\left(\frac{\varepsilon_{o} A}{d}\right) V^{2}\)

\(=\frac{\varepsilon_{o} A V^{2}}{2 d}\)

Part \(\mathrm{A}\)

The energy stored in the capacitor is \(U_{o}=\frac{\varepsilon_{o} A V^{2}}{2 d}\)

 

The energy store in the capacitor depends on the area of the plates of the capacitor, the distance between the plates, and the potential difference between the two plates.

(B) The area of the plates is also increased by a factor of 3, that is,

\(A=3 A\)

The equation for the capacitance of the capacitor when the separation is \(3 \mathrm{~d}\) is,

\(C^{\prime}=\frac{\varepsilon_{o} A}{d}\)

Substitute \(3 A\) for \(A^{\prime}\) in above equation. \(C^{\prime}=\frac{\varepsilon_{o}(3 A)}{d} \ldots \ldots\) (3)

The new energy stored in the capacitor is,

\(U_{1}=\frac{1}{2} C^{\prime} V^{2}\)

Substitute the equation (4) in above equation (3).

\(\begin{aligned} U_{1} &=\frac{1}{2}\left(\frac{\varepsilon_{o}(3 A)}{d}\right) V^{2} \\ &=\frac{3 \varepsilon_{o} A V^{2}}{2 d} \end{aligned}\)

Part B

The energy stored in the capacitor \(U_{1}=\frac{3 \varepsilon_{o} A V^{2}}{2 d}\)

Here, the energy stored in the capacitor is found after the separation of the two parallel plates is tripled, that \(3 \mathrm{~d}\).

(C) If the dielectric material is inserted between the capacitor plates, then the capacitance of the capacitor is,

\(C^{\prime \prime}=\frac{k \varepsilon_{o} A}{d} \ldots \ldots(5)\)

Here, \(\mathrm{k}\) is the electric constant. The equation for the energy stored after the dielectric material inserted between the two plates is,

\(U_{2}=\frac{1}{2} C^{\prime \prime} V^{2}\)

Substitute the equation (6) in above equation (5).

\(\begin{aligned} U_{2} &=\frac{1}{2}\left(\frac{k \varepsilon_{o} A}{d}\right) V^{2} \\ &=\frac{k \varepsilon_{o} A V^{2}}{2 d} \end{aligned}\)

Part C

The energy stored in the capacitor after the dielectric material inserted is \(U_{2}=\frac{k \varepsilon_{o} A V^{2}}{2 d}\)

Here, the dielectric material is inserted between the two plates of the capacitor, and the new energy stored in the capacitor is found.

Part A

The energy stored in the capacitor is \(U_{o}=\frac{\varepsilon_{o} A V^{2}}{2 d}\)

Part B

The energy stored in the capacitor is \(U_{1}=\frac{3 \varepsilon_{o} A V^{2}}{2 d}\)

Part C

The energy stored in the capacitor after the dielectric material inserted is \(U_{2}=\frac{k \varepsilon_{o} A V^{2}}{2 d}\)