Question

Learning Goal:

To understand that the charge stored by capacitors represents energy; to be able to calculate the stored energy and its changes under different circumstances.

An air-filled parallel-plate capacitor has plate area A and plate separation d. The capacitor is connected to a battery that creates a constant voltage V.

Part A

Find the energy U0 stored in the capacitor.

Express your answer in terms of A, d, V, and ?0.

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 ?0.

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 ?0.

Answer 1

Concepts and reason

The concepts used to solve this problem are capacitance of a capacitor and the energy stored by the capacitor.

Use the expression of capacitance of a parallel plate capacitor and then calculate the energy stored by the capacitor.

Use the concept that when capacitor is disconnected from the battery, the charge on it remains constant. On changing the distance between the plates the capacitance and voltage of the capacitor change. Use these new parameters of the capacitor to calculate the energy stored in the capacitor.

Use the concept that inserting a dielectric between the plates of the capacitor changes the capacitance. Use the new parameters to calculate the energy stored in the capacitor.

Fundamentals

Capacitance is the ability of the capacitor to store electric charge. A capacitor is a device that stores electric charge and electrostatic energy.

The capacitance $C$ of a capacitor is given by the relation:

$C = \frac{Q}{V}$

Here, $Q$ is the magnitude of the electric charge stored by the capacitor having a potential difference of $V$volts.

The S.I Units of capacitance is farad (F).

The capacitance of a parallel-plate capacitor is given by:

$C = \frac{{A{\varepsilon _ \circ }}}{d}$

Here,$A$is the area of cross section of capacitor, $d$ is distance between the plates and ${\varepsilon _ \circ }$ is the permittivity of free space.

If a dielectric slab with dielectric constant $k$ is inserted between the capacitor, the new capacitance of the parallel capacitor is given by the expression:

$C = \frac{{Ak{\varepsilon _ \circ }}}{d}$

The energy stored by a capacitor is given by the relation:

$U = \frac{1}{2}C{V^2}$

Here,$C$ is the capacitance and $V$ is the potential difference.

(a)

Write the expression for capacitance of a parallel-plate capacitor.

$C = \frac{{A{\varepsilon _ \circ }}}{d}$

Write the expression for energy stored by a capacitor.

$U = \frac{1}{2}C{V^2}$

Substitute $\frac{{A{\varepsilon _ \circ }}}{d}$ for $C$ and ${U_ \circ }$ for U.

$\begin{array}{c}\\{U_ \circ } = \frac{1}{2}\left( {\frac{{A{\varepsilon _ \circ }}}{d}} \right){V^2}\\\\ = \frac{{{\varepsilon _ \circ }A{V^2}}}{{2d}}\\\end{array}$

(b)

Write the expression of charge stored in the capacitor.

$C = \frac{Q}{V}$

Rearrange the expression for change in the capacitor.

$Q = CV$

Substitute $\frac{{A{\varepsilon _ \circ }}}{d}$ for $C$.

$Q = \frac{{A{\varepsilon _ \circ }V}}{d}$

Since, the battery is disconnected the amount of charge on the capacitor will remain same.

The new distance between the plates will change the capacitance and voltage and the charge remains constant.

$\begin{array}{l}\\C' = \frac{{A{\varepsilon _ \circ }}}{{3d}}\\\\C' = \frac{C}{3}\\\end{array}$

Here, $C'$ is the new capacitance.

The voltage increases by the same factor by which the capacitance is decreased.

$V' = 3V$

Here, $V'$is the new potential difference.

Write the expression for new energy ${U_1}$.

${U_1} = \frac{1}{2}C'{V'^2}$

Substitute $\frac{{A{\varepsilon _ \circ }}}{{3d}}$for $C'$ and $3V$for $V'$.

$\begin{array}{c}\\{U_1} = \frac{1}{2}\frac{{A{\varepsilon _ \circ }}}{{3d}}\left( {3V} \right)\left( {3V} \right)\\\\ = \frac{3}{2}\frac{{{\varepsilon _ \circ }A{V^2}}}{d}\\\end{array}$

(c)

The distance between the plates is restored to $d$.

Write the expression of capacitance.

$C = \frac{{A{\varepsilon _ \circ }}}{d}$

Now, the capacitor is reconnected to the battery. Hence, the capacitor has a constant voltage.

Write the expression of capacitance after the dielectric is inserted between the plates of capacitor.

$C = \frac{{Ak{\varepsilon _ \circ }}}{d}$

Write the expression of energy stored in the capacitor.

$\begin{array}{c}\\{U_2} = \frac{1}{2}C{V^2}\\\\ = \frac{1}{2}\left( {\frac{{Ak{\varepsilon _ \circ }}}{d}} \right){V^2}\\\\ = \frac{{kA{\varepsilon _ \circ }{V^2}}}{{2d}}\\\end{array}$

Ans: Part a

Energy stored in the capacitor ${U_ \circ } = \frac{{{\varepsilon _ \circ }A{V^2}}}{{2d}}$ .

Part b

Energy stored in the capacitor ${U_1} = \frac{3}{2}\frac{{{\varepsilon _ \circ }A{V^2}}}{d}$

Part c

Energy stored in the capacitor ${U_2} = \frac{{kA{\varepsilon _ \circ }{V^2}}}{{2d}}$