The parallel combination of two air filled parallel plate capacitors of capacitance C and nC is conn

Q: Why does the total charge remain constant after removing the battery?

Because the charge has nowhere to go! The circuit is broken at the battery terminals, leaving the plates isolated. By the law of conservation of charge, the total trapped charge must remain constant.

Q: Why does the capacitance increase when a dielectric is inserted?

A dielectric material gets polarized in the electric field, which partially cancels the internal field. This allows the plates to store more charge for the same potential difference, effectively increasing the capacitance.

Q: Does the charge on each individual capacitor remain the same after inserting the dielectric?

No! While the total charge is conserved, the charge redistributes between the two capacitors because their relative capacitances have changed. The capacitor with the dielectric will draw more charge to maintain the same potential difference across both.

Q: Why are the capacitors still in parallel after the battery is removed?

Their top plates are still connected to each other by a conducting wire, and their bottom plates are connected to each other. This means they must always share the same potential difference, which is the definition of a parallel combination.

Q: What would happen if the battery was kept connected while inserting the dielectric?

If the battery remained connected, the potential difference would be forced to stay constant at $V$. The equivalent capacitance would still increase, meaning the battery would have to supply extra charge to the system.

Question

The parallel combination of two air filled parallel plate capacitors of capacitance $C$ and $nC$ is connected to a battery of voltage, $V$. When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant $K$ is placed between the two plates of the first capacitor. The new potential difference of the combined system is

Options:

(A) $\frac{(n+1)V}{(K+n)}$

(B) $\frac{nV}{K+n}$

(C) $V$

(D) $\frac{V}{K+n}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a simple yet elegant circuit. We have two air-filled parallel plate capacitors, one with capacitance $C$ and the other with capacitance $nC$. They are connected in parallel to a battery that provides a voltage $V$.

Because they are in parallel, their equivalent capacitance is simply the sum of their individual capacitances. So, we can write:

$$C_{eq} = C + nC = C(1+n)$$

The battery does its job and fully charges this combination. The total charge $Q$ supplied by the battery is the product of the equivalent capacitance and the voltage. Therefore:

$$Q = C_{eq}V = C(1+n)V$$

### The Master Equation

Now, we disconnect the battery. This is a crucial moment! By removing the battery, we cut off the path for any charge to enter or leave the capacitors. **The total charge $Q$ is now trapped and must remain strictly conserved.**

Next, we introduce a twist. We slide a dielectric material with a dielectric constant $K$ between the plates of the first capacitor. This action increases its capacitance by a factor of $K$, making its new capacitance $KC$.

With this change, the new equivalent capacitance of our parallel system becomes:

$$C'_{eq} = KC + nC = C(K+n)$$

### Final Calculation

Even though the capacitance has changed, the total charge is still trapped. **By the law of conservation of charge, the final total charge must equal the initial total charge.** We can express this as $Q_{final} = Q_{initial}$.

Let the new potential difference across the combination be $V'$. The final charge can be written as $C'_{eq}V'$. Equating this to our initial charge, we get:

$$C'_{eq}V' = C_{eq}V$$

Substituting the expressions we found earlier, we have:

$$C(K+n)V' = C(1+n)V$$

Notice how the original capacitance $C$ beautifully cancels out from both sides. Solving for the new potential difference, we arrive at our final answer:

$$V' = \frac{(1+n)V}{K+n}$$

It is a brilliant demonstration of how charge conservation governs the behavior of isolated systems.

Select Answer:

Visualized Solution

Initial Setup

Initial Equivalent Capacitance \& Charge

Battery Removed

Dielectric Insertion

New Equivalent Capacitance

Conservation of Charge

Substituting Values

Final Potential Difference

The Sigma Insight: Combination of Capacitors

Analyzing the Setup

The Master Equation

Final Calculation

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