A glass slab is introduced between the plates of a parallel plate capacitor. Does the capacitance of the capac

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Q. A glass slab is introduced between the plates of a parallel plate capacitor. Does the capacitance of the capacitor increase, decrease or remain unchanged? Two capacitors of capacitances 5 μF and 10 μF are charged to 16 volts and 10 volts respectively. Find the common potential when they are connected in parallel. OR 64 identical water drops coalesce to form a larger drop. If the nature and amount of charge be the same for all the drops, calculate the potential, capacitance and stored energy of the larger drop.

West Bengal WbchseWest Bengal HS (WBCHSE) Board 2016Subjective· 3mImportance★★★★★

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Concept understanding — Capacitor Network Analysis

Capacitor Network Analysis

Imagine you have a bucket of water and a pipe. The bigger the bucket, the more water it can hold for a given water pressure. A capacitor does the same thing with electric charge — it stores charge when a voltage is applied. The "size" of the bucket is called capacitance ( $C$ ), measured in farads (F).

Now, what happens when you connect several buckets together with pipes? That's a capacitor network. The analysis is about finding one equivalent bucket (a single capacitor) that behaves exactly like the whole network.

The Core Idea: Charge and Voltage Must Match

When you connect capacitors, two things are always true:

Charge conservation: Total charge in a closed part of the circuit stays the same unless a battery pushes more in.
Voltage is shared: The voltage across each capacitor depends on how they're wired.

There are only two basic ways to connect them. Everything else is a combination of these two.

Series Connection: One Path, Shared Charge

Connect capacitors end-to-end, like train cars. The same current flows through each, so each capacitor stores the same amount of charge $Q$ .

But the total voltage across the combination is the sum of the individual voltages:

V_{\text{total}} = V_1 + V_2 + V_3 + \dots

Since $V = Q/C$ for each capacitor, we get:

\frac{Q}{C_{\text{eq}}} = \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3} + \dots

Cancel $Q$ (it's the same everywhere):

\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \dots

Intuition: Adding capacitors in series makes the equivalent capacitance smaller than the smallest individual one. Why? Because you're effectively making the "bucket" longer and narrower — harder to fill.

Watch out

A common mistake: students treat series capacitors like series resistors (adding reciprocals for resistors, but adding directly for capacitors). It's the opposite. For resistors in series: $R_{\text{eq}} = R_1 + R_2$ . For capacitors in series: $1/C_{\text{eq}} = 1/C_1 + 1/C_2$ .

Parallel Connection: Multiple Paths, Same Voltage

Connect capacitors side by side, like buckets with their bottoms connected by a wide pipe. Each capacitor sees the same voltage $V$ across it.

But the total charge stored is the sum of charges on each:

Q_{\text{total}} = Q_1 + Q_2 + Q_3 + \dots

Since $Q = CV$ for each:

C_{\text{eq}} V = C_1 V + C_2 V + C_3 V + \dots

Cancel $V$ :

C_{\text{eq}} = C_1 + C_2 + C_3 + \dots

Intuition: Adding capacitors in parallel makes the equivalent capacitance larger — you're just adding more bucket area. Easy to fill.

Tip

| Connection | Equivalent Formula | What happens to $C_{\text{eq}}$ |

|------------|-------------------|----------------------------------|

| Series | $1/C_{\text{eq}} = \sum 1/C_i$ | Gets smaller than smallest |

| Parallel | $C_{\text{eq}} = \sum C_i$ | Gets larger than largest |

How to Analyze Any Network

Spot the pattern: Look for capacitors that are clearly in series (only two terminals, no branching between them) or clearly in parallel (both ends connected together).
Replace step by step: Replace each simple series or parallel group with its equivalent capacitor. Redraw the circuit after each step.
Repeat until you have one capacitor.

Note

This is exactly like simplifying resistor networks — but with the reciprocal formula for series. If you can do resistor networks, you can do capacitor networks. Just flip the series formula.

A Worked Example

Suppose you have three capacitors: $C_1 = 2\ \mu\text{F}$ , $C_2 = 3\ \mu\text{F}$ , $C_3 = 6\ \mu\text{F}$ . $C_2$ and $C_3$ are in parallel, and that combination is in series with $C_1$ . …

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Inserting a dielectric (glass) always increases a capacitor's capacitance since C = Kε₀A/d and K(glass) > 1; and when charged capacitors are joined in parallel, charge redistributes until they share one common potential, here 12 V.

Part 1 — Effect of glass slab: The capacitance of a parallel plate capacitor with a dielectric of dielectric constant K filling the gap is C = Kε₀A/d, compared to C₀ = ε₀A/d without it. Since glass has K > 1 (typically 5-10), inserting the slab INCREASES the capacitance.

Part 2 — Common potential: When two charged capacitors are connected in parallel, charge is conserved and redistributes to give a common potential equal to the charge-weighted average:

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set SEM31 markMCQ
$Q. Three capacitors having capacitances 1\cdot0 μF, 2\cdot0 μF and 5\cdot0 μF are connected in series with a source of 10 V. The potential difference between the two ends of the capacitor having capacitance 2\cdot0 μF will be$
(a) 100 V / 17
(b) 20 V / 17
(c) 50 V / 17
(d) 10 V
›Reveal solutionSolution

In series, all capacitors carry the same charge. C_eq = 10/17 μF, Q = 100/17 μC, and V₂ = Q/(2 μF) = 50/17 V. Option (c).

Step 1 — equivalent series capacitance:

1/C_eq = 1/1 + 1/2 + 1/5 = (10 + 5 + 2)/10 = 17/10, so C_eq = 10/17 μF.

Step 2 — common charge (series capacitors share the same charge):

Q = C_eq × V = (10/17 μF)(10 V) = 100/17 μC.

Step 3 — potential difference across the 2·0 μF capacitor: …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. When two identical capacitors are in series, they have 3 uF resultant capacitance and when parallel 12 uF. What is the capacitance of each?$
(i) 6 uF
(ii) 3 uF
(iii) 12 uF
(iv) 9 uF
›Reveal solutionSolution

Each capacitor is 6 uF.

…

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CBSE 2024Set 55/1/11 markMCQ
$1\ \mu\text{F}$
›Reveal solutionSolution
For capacitors in parallel, the total capacitance is the sum of individual capacitances. Here, $C_{\text{eq}} = 10\ \mu\text{F}$ . Energy stored is $\frac{1}{2} C_{\text{eq}} V^2 = \frac{1}{2} \times 10 \times 10^{-6} \times (100)^2 = 0.05\ \text{J} = 5.0 \times 10^{-2}\ \text{J}$ . The correct option is (D).

When capacitors are connected in parallel, the voltage across each capacitor is the same — equal to the source voltage. This is the key difference from series connections, where the charge is the same but voltage divides. Because all ten capacitors are identical and each sees the full $100\ \text{V}$ , the total energy is simply the sum of the energies stored in each capacitor individually.

The energy stored in a single capacitor of capacitance $C$ at voltage $V$ is $\frac{1}{2} C V^2$ . For ten such capacitors, the total energy is $10 \times \frac{1}{2} C V^2 = \frac{1}{2} (10C) V^2$ . Notice that $10C$ is exactly the equivalent capacitance of ten $1\ \mu\text{F}$ capacitors in parallel. So the problem reduces to finding the energy stored in a single $10\ \mu\text{F}$ capacitor charged to $100\ \text{V}$ .

Let’s work through the numbers carefully.

Find the equivalent capacitance.

For parallel combination: $C_{\text{eq}} = C_1 + C_2 + \dots + C_{10} = 10 \times 1\ \mu\text{F} = 10\ \mu\text{F}$ .

In SI units: $10\ \mu\text{F} = 10 \times 10^{-6}\ \text{F} = 10^{-5}\ \text{F}$ .

Apply the energy formula.

The energy stored in a capacitor network (or a single equivalent capacitor) is $U = \frac{1}{2} C_{\text{eq}} V^2$ , where $V$ is the voltage across the combination.

Here $V = 100\ \text{V}$ , so:

$U = \frac{1}{2} \times (10^{-5}) \times (100)^2 = \frac{1}{2} \times 10^{-5} \times 10^4 = \frac{1}{2} \times 10^{-1} = 0.05\ \text{J}.$ …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. Three capacitors are connected in triangle as shown in figure. The equivalent capacitance between the points A and C is :$
(a) 4 $\mu$ F
(b) 2 $\mu$ F
(c) 8 $\mu$ F
(d) 6 $\mu$ F
›Reveal solutionSolution
The direct A-C capacitor is in parallel with the series A-B-C path (which reduces to $2\,\mu F$ ), giving a total equivalent capacitance of $6\,\mu F$ between A and C.

Working

Three $4\,\mu F$ capacitors form a triangle with vertices A, B, C: one directly between A and C, one between A and B, and one between B and C.

Between the terminals A and C, there are two parallel paths:

The capacitor directly connecting A to C: $C_1 = 4\,\mu F$ .

The path A $\to$ B $\to$ C, formed by the A-B and B-C capacitors in series: …
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CBSE 2023Set ANNUAL1 markMCQ
$Q. Three capacitors each of capacitance 4 μF are to be connected in such a way that the effective capacitance is 6 μF. This can be done by connecting$
(a) each of them in series
(b) each of them in parallel
(c) two in parallel and one in series
(d) two in series and one in parallel
›Reveal solutionSolution

Two 4 μF capacitors in series (2 μF) placed in parallel with the third 4 μF capacitor gives 2 + 4 = 6 μF.

Each capacitor has $C = 4\ \mu F$ . Check the given option: two in series, then that combination in parallel with the third.

Series combination of two $4\ \mu F$ capacitors:

$\dfrac{1}{C_s} = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2} \implies C_s = 2\ \mu F$

This series pair is now connected in parallel with the third $4\ \mu F$ capacitor:

$C_{eff} = C_s + 4 = 2 + 4 = 6\ \mu F$

…

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CBSE 2022Set ANNUAL1 markMCQ
$Q. Three capacitors of equal capacity C are joined first in parallel and then in series. The ratio of equivalent capacities in both the cases will be:$
(a) 9 : 1
(b) 6 : 1
(c) 3 : 1
(d) 1 : 9
›Reveal solutionSolution

$C_{parallel} = 3C$ and $C_{series} = C/3$ , so the ratio is $9 : 1$ : option (A).

For three equal capacitors each of capacitance $C$ :

In parallel, capacitances add:

$C_p = C + C + C = 3C.$

In series, reciprocals add: …

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CBSE 2020Set 55/3/11 markMCQ
$C_1$
›Reveal solutionSolution
In a parallel combination, both capacitors share the same voltage. Since $Q = CV$ , the charge on each is directly proportional to its capacitance, so the ratio $Q_1 : Q_2 = C_1 : C_2$ . The answer is $\boxed{\dfrac{C_1}{C_2}}$ , option (A).

When two capacitors are connected in parallel, the defining feature is that they are connected across the same two points. This means the potential difference (voltage) across each capacitor is identical. This is the single most important idea — everything else follows from it.

Think of it like two buckets of different widths placed side by side under the same tap. The water level (voltage) in both will be the same, but the wider bucket (larger capacitance) will hold more water (charge). The charge stored by a capacitor is $Q = CV$ , so if $V$ is fixed, the charge is simply proportional to $C$ .

Now, let’s work through it step by step.

State the parallel condition.

For capacitors $C_1$ and $C_2$ in parallel, the voltage across each is the same. Call this common voltage $V$ .

Write the charge on each capacitor.

Using $Q = CV$ :

$Q_1 = C_1 V \quad \text{and} \quad Q_2 = C_2 V$

Find the required ratio. The ratio of the charge on $C_1$ to the charge on $C_2$ is:

$\frac{Q_1}{Q_2} = \frac{C_1 V}{C_2 V} = \frac{C_1}{C_2}$
The voltage $V$ cancels out completely — the ratio depends only on the capacitances.

Match with the options. The ratio $\dfrac{C_1}{C_2}$ corresponds directly to option (A). …
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CBSE 2018Set ANNUAL1 markMCQ
$Q. Two capacitors C₁ = 2μF and C₂ = 4μF are connected in series and a potential difference (p.d.) of 1200V is applied across it. The Potential difference across 2μF will be :-$
(a) 400V
(b) 600V
(c) 800V
(d) 900V
›Reveal solutionSolution

Series capacitors share equal charge, so V is inversely proportional to C; the 2μF takes 800 V.

Series combination: same charge Q on both plates.

$Q = C_1 V_1 = C_2 V_2$ , so $\dfrac{V_1}{V_2} = \dfrac{C_2}{C_1} = \dfrac{4}{2} = 2.$

Thus $V_1 = 2V_2$ (V₁ across the 2μF). With $V_1 + V_2 = 1200$ V:

$2V_2 + V_2 = 1200 \Rightarrow 3V_2 = 1200 \Rightarrow V_2 = 400$ V (across 4μF), and $V_1 = 800$ V (across 2μF).

…

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CBSE 2018Set ANNUAL1 markMCQ
$Q. Minimum number of capacitors of 2μF each required to obtain a capacitance of 5μF will be :-$
(a) 4
(b) 3
(c) 5
(d) 6
›Reveal solutionSolution
5μF = (two 2μF in parallel = 4μF) in parallel with (two 2μF in series = 1μF) → 4 capacitors.

We want 5μF from 2μF units.

Two 2μF in series: $\dfrac{2\times2}{2+2}=1\,\mu F$ .

Two 2μF in parallel: $2+2=4\,\mu F$ .

Put the 4μF branch in parallel with the 1μF branch: $4+1 = 5\,\mu F$ . …
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CBSE 2016Set ANNUAL1 markMCQ
$Q. A capacitor of capacitance C1 is charged up to potential V and then connected in parallel to an uncharged capacitor of capacitance C2. The final potential difference across each capacitor will be$
(a) C2V/(C1+C2)
(b) C1V/(C1+C2)
(c) (1+C2/C1)V
(d) (1-C2/C1)V
›Reveal solutionSolution

Charge is conserved when the charged capacitor is connected to the uncharged one; sharing it between the two parallel capacitances gives the common final potential.

…

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