The figure shows two identical parallel plate capacitors A and B connected to a battery with the swi

Q: Why does the voltage across capacitor A remain constant?

Because capacitor A is directly connected to the terminals of the battery. As long as the battery is connected, it will supply or draw charge to maintain the potential difference $V$ across the capacitor plates.

Q: Why does the charge on capacitor B remain constant after opening the switch?

Once the switch is opened, the circuit branch containing capacitor B is broken. There is no conducting path for the electrons to leave or enter the plates of capacitor B. Therefore, the charge it acquired initially is trapped and must remain constant.

Q: Why do we use different formulas for the final energy of A and B?

Energy can be calculated using $U = \frac{1}{2}CV^2$ or $U = \frac{q^2}{2C}$. For capacitor A, the voltage $V$ is constant, so using $\frac{1}{2}CV^2$ is straightforward. For capacitor B, the charge $q$ is constant, so using $\frac{q^2}{2C}$ avoids the need to calculate the new, altered voltage across B.

Q: What happens to the voltage across capacitor B after the dielectric is inserted?

Since $q$ is constant and $C$ increases to $3C$, the new voltage $V' = \frac{q}{3C} = \frac{V}{3}$. The voltage drops to one-third of its original value!

Q: Is energy conserved in this process?

No, the total electrostatic energy changes from $CV^2$ to $\frac{5}{3}CV^2$. The battery does work to pump more charge into capacitor A, and external work is done (or extracted) when inserting the dielectric slabs.

Question

The figure shows two identical parallel plate capacitors $A$ and $B$ connected to a battery with the switch $S$ closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant (or relative permittivity) 3. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a beautifully symmetric circuit. We have two identical parallel plate capacitors, $A$ and $B$, connected in parallel to a steady battery of voltage $V$.

The switch $S$ is initially closed, meaning both capacitors are enjoying the full potential difference provided by the battery.

This is our starting point, a state of perfect electrical harmony.

### The Initial Energy: Before the Split

Since both capacitors are directly connected across the battery, the potential difference across each is exactly $V$.

The energy stored in a capacitor is given by the classic formula $U = \frac{1}{2}CV^2$.

Because we have two identical capacitors, the total initial energy of the system is simply the sum of their individual energies.

$$U_i = U_A + U_B = \frac{1}{2}CV^2 + \frac{1}{2}CV^2 = CV^2$$

This $CV^2$ is our baseline energy. Keep this value in mind as we introduce some chaos into the system!

### The Great Divide: Opening the Switch

Now, we open the switch $S$. This is where the physics gets incredibly interesting!

By opening the switch, we break the circuit branch containing capacitor $B$. Capacitor $A$ remains happily connected to the battery, so its voltage is firmly locked at $V$.

However, capacitor $B$ is now completely isolated from the rest of the world. The charge it acquired initially, which is $q_B = CV$, is now trapped on its plates. It has nowhere to go!

**This is a crucial principle:** A connected capacitor maintains constant voltage, while an isolated capacitor maintains constant charge.

### The Dielectric Invasion

Next, we fill the free space of both capacitors with a dielectric material of constant $K = 3$.

As we know from the principles of electrostatics, introducing a dielectric increases the capacitance by a factor of $K$.

So, the new capacitance for both $A$ and $B$ becomes $3C$.

$$C_A' = 3C$$
$$C_B' = 3C$$

But how does this affect their energies? Let's find out!

### The Final Energy: A Tale of Two States

Let's calculate the new energy for capacitor $A$. Because it's still connected to the battery, its voltage is still $V$.

Using the formula $U = \frac{1}{2}CV^2$ with the new capacitance $3C$, its final energy becomes:

$$U_{Af} = \frac{1}{2}(3C)V^2 = \frac{3}{2}CV^2$$

For capacitor $B$, the situation is entirely different. It's isolated, so its charge $q$ remains constant. To find its energy, we must use the formula $U = \frac{q^2}{2C}$.

Substituting the trapped charge $CV$ and the new capacitance $3C$, the final energy of $B$ becomes:

$$U_{Bf} = \frac{(CV)^2}{2(3C)} = \frac{C^2V^2}{6C} = \frac{1}{6}CV^2$$

Now, we just add the final energies of $A$ and $B$ together to get the total final energy.

$$U_f = U_{Af} + U_{Bf} = \frac{3}{2}CV^2 + \frac{1}{6}CV^2$$

Finding a common denominator, we get:

$$U_f = \frac{9}{6}CV^2 + \frac{1}{6}CV^2 = \frac{10}{6}CV^2 = \frac{5}{3}CV^2$$

### The Grand Finale: The Ratio

Finally, we find the ratio of the initial energy to the final energy.

$$	ext{Ratio} = \frac{U_i}{U_f} = \frac{CV^2}{\frac{5}{3}CV^2}$$

The $CV^2$ terms cancel out completely, leaving us with the final ratio:

$$	ext{Ratio} = \frac{3}{5}$$

And there we have it! A perfect, elegant result derived from understanding the fundamental difference between connected and isolated capacitors.

Visualized Solution

$Initial Setup$

$Initial Energy$

$Opening the Switch$

$Inserting Dielectric$

$Final Energy of A$

$Final Energy of B$

$Total Final Energy$

$Ratio of Energies$

The Sigma Insight: Capacitance and Capacitors

Analyzing the Setup

The Initial Energy

The Great Divide

The Dielectric Invasion

The Final Energy

The Grand Finale

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Opening the Switch

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Inserting Dielectric

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Energy of A

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Energy of B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Final Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ratio of Energies