Two parallel plate capacitors of capacitances C and 2C are connected in parallel and charged to a po

Q: Why does the total charge remain constant?

When the battery is disconnected, the circuit is broken. The electrons have nowhere to go, so the total charge trapped on the isolated plates of the parallel combination must remain exactly the same.

Q: Why does the capacitance of the first capacitor become $KC$?

A dielectric material polarizes in the presence of an electric field, effectively reducing the internal electric field and allowing the capacitor to store more charge for the same voltage. This increases its capacitance by a factor of its dielectric constant $K$.

Q: What would happen if the battery remained connected?

If the battery stayed connected, the potential difference across the capacitors would be forced to remain $V$. Inserting the dielectric would then cause the battery to supply more charge to the system.

Q: Does the charge on the $2C$ capacitor change?

Yes! Even though the *total* charge is conserved, the insertion of the dielectric changes the relative capacities. Charge will redistribute, flowing from the $2C$ capacitor to the $KC$ capacitor until they both reach the new common potential $V'$.

Q: Is energy conserved in this process?

No, the total electrostatic potential energy decreases. The electric field of the capacitors actually pulls the dielectric slab inwards, doing positive work. This work comes at the expense of the stored electrostatic energy.

Question

Two parallel plate capacitors of capacitances $C$ and $2C$ are connected in parallel and charged to a potential difference $V$. The battery is then disconnected and the region between the plates of capacitor $C$ is completely filled with a material of dielectric constant $K$. The potential difference across the capacitors now becomes....

Sigma Percentile · Accepted Answer

### The Tale of the Trapped Charge: Dielectrics in Parallel Capacitors

Imagine you have a system of two capacitors, $C$ and $2C$, connected in parallel across a battery of voltage $V$. This is our starting point. When capacitors are in parallel, their equivalent capacitance is simply the sum of their individual capacitances.

So, initially, the equivalent capacitance is:
$$C_{eq} = C + 2C = 3C$$

The battery acts as a pump, pushing charge onto the plates until the potential difference across the capacitors matches the battery's voltage $V$. The total charge $Q_{initial}$ supplied to the system is:
$$Q_{initial} = C_{eq}V = 3CV$$

### The Disconnection and the Dielectric

Now comes the crucial twist: the battery is disconnected. This breaks the circuit, meaning the electrons have nowhere to go. The total charge $3CV$ is now permanently trapped on the isolated plates of our parallel combination. This is the principle of **conservation of charge**.

Next, a dielectric material with a dielectric constant $K$ is completely filled into the region between the plates of the first capacitor $C$. A dielectric polarizes in the presence of an electric field, which effectively increases the capacitance of that specific capacitor by a factor of $K$.

The new capacitance of the first capacitor becomes $KC$. The second capacitor remains untouched at $2C$. Since they are still connected in parallel, our new equivalent capacitance is:
$$C'_{eq} = KC + 2C = (K+2)C$$

### Finding the New Voltage

Because the total charge is trapped, the final total charge $Q_{final}$ must equal the initial total charge $Q_{initial}$. Let the new potential difference across the combination be $V'$. We can express the final charge as:
$$Q_{final} = C'_{eq}V'$$

Equating the initial and final charges:
$$(K+2)C \cdot V' = 3CV$$

Notice how beautifully the $C$ cancels out from both sides. Solving for the new potential difference $V'$, we get:
$$V' = \frac{3V}{K+2}$$

### The Energy Perspective

It is fascinating to think about what happens to the stored electrostatic energy during this process. The energy of a capacitor can be expressed as $U = \frac{Q^2}{2C_{eq}}$.

Since the total charge $Q$ remained strictly constant while the equivalent capacitance $C_{eq}$ increased (from $3C$ to $(K+2)C$), the total stored energy must have **decreased**. But energy cannot just disappear! Where did it go?

The electric field between the plates of the capacitor actually exerts an attractive force on the dielectric slab, pulling it inwards. The capacitor does positive mechanical work on the slab, and this work comes directly at the expense of its stored electrostatic potential energy.

Visualized Solution

$C_{e q}$

$Q_{ini t ia l}$

$Q_{f ina l} = Q_{ini t ia l}$

$C_{e q}^{'}$

$V^{'}$

$C_{e q}^{'} V^{'} = Q_{ini t ia l}$

$V^{'} = \frac{3}{K + 2} V$

$U = \frac{Q ^{2}}{2 C _{e q}}$

The Sigma Insight: Combination of Capacitors

The Tale of the Trapped Charge

The Disconnection and the Dielectric

Finding the New Voltage

The Energy Perspective

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Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Qfinal​=Qinitial​

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ceq′​V′=Qinitial​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V′=K+23​V

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } U=2Ceq​Q2​