Two capacitors of capacities 2C and C are joined in parallel and charged upto potential V. The batte

Q: Why does the total charge remain constant?

Because the battery is removed, there is no external circuit for the charge to flow into or out of. The charge is trapped on the plates of the capacitors.

Q: Why do we add the capacitances?

The capacitors are connected in parallel. In a parallel combination, the equivalent capacitance is the algebraic sum of the individual capacitances.

Q: Does the potential difference change when the dielectric is inserted?

Yes. Since the charge is constant and the equivalent capacitance increases, the potential difference must decrease according to $V = Q/C$.

Q: What happens to the energy of the system?

The energy decreases. The work done by the electric field to pull the dielectric into the capacitor accounts for this loss in stored electrostatic energy.

Q: Why doesn't the charge on each individual capacitor remain the same?

While the *total* charge is conserved, charge will redistribute between the two capacitors to ensure they both have the same new potential difference, as they are in parallel.

Question

Two capacitors of capacities $2C$ and $C$ are joined in parallel and charged upto potential $V$. The battery is removed and the capacitor of capacity $C$ is filled completely with a medium of dielectric constant $K$. The potential difference across the capacitors will now be

Options:

(A) $\frac{V}{K+2}$

(B) $\frac{V}{K}$

(C) $\frac{3V}{K+2}$

(D) $\frac{3V}{K}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you have two buckets of water (our capacitors) connected by a pipe (in parallel). One bucket is twice as large as the other ($2C$ and $C$). You fill them up using a hose (the battery) until the water level in both reaches a certain height ($V$).

Because they are connected in parallel, they share the same potential difference. The total amount of water (charge) you've poured in is the sum of the water in each bucket.

$$Q_{initial} = Q_1 + Q_2 = (2C)V + (C)V = 3CV$$

### The Transformation

Now, you completely remove the hose. The total amount of water in the buckets is now trapped. It cannot escape. This is a crucial principle: **when a battery is disconnected, the total charge on the isolated system remains strictly conserved.**

Next, you drop a solid block (the dielectric) into the smaller bucket. What does this do? In the world of capacitors, inserting a dielectric of constant $K$ increases the bucket's capacity to hold charge by a factor of $K$.

So, the new capacitance of the second capacitor becomes $KC$. The first capacitor is untouched, so it remains $2C$.

### The Master Equation

Even though the capacities have changed, the two buckets are still connected by the pipe. They must still share the same water level (potential difference).

To find this new level, we first need the new total capacity of our system. Since they are in parallel, we simply add them up:

$$C_{eq}' = 2C + KC = C(K + 2)$$

### Final Calculation

We know the total trapped charge is $3CV$, and the new total capacity is $C(K + 2)$. The new potential difference $V'$ is simply the total charge divided by the total capacity:

$$V' = \frac{Q_{final}}{C_{eq}'}$$

Substituting our values:

$$V' = \frac{3CV}{C(K + 2)}$$

The $C$ beautifully cancels out from the numerator and denominator, leaving us with our final, elegant result:

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

This tells us exactly how the voltage drops when the dielectric is introduced, perfectly balancing the increased capacity with the conserved charge.

Select Answer:

Visualized Solution

$I ni t ia l S e t u p$

$T o t a l I ni t ia l C ha r g e$

$C a l c u l a t in g Q$

$B a tt er y R e m o v e d$

$N e w C a p a c i t an ces$

$E q u i v a l e n tC a p a c i t an ce$

$F ina l P o t e n t ia l$

The Sigma Insight: Combination of Capacitors

Analyzing the Setup

The Transformation

The Master Equation

Final Calculation

Similar Questions

A $10 μ F$ capacitor is fully charged to a potential difference of 50 V. After removing the source voltage it is connected to an uncharged capacitor in parallel. Now, the potential difference across them becomes 20 V. The capacitance of the second capacitor is

A capacitor $C$ is fully charged with voltage $V_{0}$ . After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance $\frac{C}{2}$ . The energy loss in the process after the charge is distributed between the two capacitors is

Two capacitors of capacitances $C$ and $2 C$ are charged to potential differences $V$ and $2 V$ respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is

Two capacitors $C_{1}$ and $C_{2}$ are charged to $120 V$ and $200 V$ , respectively. It is found that by connecting them together, the potential on each one can be made zero. Then,

A $60 pF$ capacitor is fully charged by a $20 V$ supply. It is then disconnected from the supply and is connected to another uncharged $60 pF$ capacitor in parallel. The electrostatic energy that is lost in this process by the time, the charge is redistributed between them is (in $nJ$ )

Select Answer:

Visualized Solution

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

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

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

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

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

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

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