Capacitance

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Home Concept Boosters CBSE Class 12 Physics Capacitance – Concept Booster | Class 12 Physics CBSE

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How to Use This Page

Read each concept carefully, then check the formula, common mistake, and exam tip before moving to the next. This page covers Capacitance completely for CBSE Class 12 Physics.

Key Concepts

Class 12 · Physics · Electrostatic Potential and Capacitance

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Core concepts you must know

Class 12 · Ch 2

Electrical Capacitance Definition

The measure of a conductor’s ability to store electric charge and potential energy. It is the ratio of charge given to the conductor to the potential developed. SI Unit: Farad (F). Practical Unit: Microfarad ($\mu$F).

$$C = \frac{q}{V}$$

Capacitance of an Isolated Sphere Formula

The capacitance of a spherical conductor depends directly on its radius $r$ and the permittivity of the surrounding medium. It shows that larger objects can store more charge at the same potential.

$$C = 4 \pi \varepsilon_0 r$$

Capacitance of a Parallel Plate Capacitor Formula

Consists of two large parallel conducting plates of area $A$, separated by a distance $d$ in a vacuum/air. Capacitance increases with larger plates and smaller separation gaps.

$$C = \frac{A \varepsilon_0}{d}$$

Capacitors in Series Formula

When capacitors are connected end-to-end, the charge on each is the same, but the total potential difference is the sum of individual potentials. The equivalent capacitance is always less than the smallest individual capacitance.

$$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$$

Capacitors in Parallel Formula

When capacitors are connected across the same two points, the potential difference across each is the same, but the total charge is the sum of individual charges.

$$C = C_1 + C_2 + C_3$$

Capacitor with Dielectric Slab Formula

Inserting a dielectric (insulating) slab of thickness $t$ and dielectric constant $K_D$ partially between the plates reduces the net electric field, lowering potential and thereby increasing capacitance.

$$C = \frac{\varepsilon_0 A}{d – t\left(1 – \frac{1}{K_D}\right)}$$

Capacitor with Conducting Slab Formula

If a metal/conducting slab of thickness $t$ is placed between the plates, the electric field inside the metal becomes zero, effectively shrinking the gap between the plates to $(d-t)$.

$$C = \frac{C_0}{\left(1 – \frac{t}{d}\right)}$$

Energy Stored in Capacitor Formula

The work done in transferring charges from one plate to the other against the opposing electric field is stored as electrostatic potential energy ($U$) in Joules (J).

$$U = \frac{q^2}{2 C} = \frac{1}{2} C V^2 = \frac{1}{2} q V$$

Concept Deep Dive

Dielectric vs. Conducting Slabs

How intervening mediums change capacity

Core Concept

When a conducting slab is placed between the plates, free electrons inside it instantly rearrange to completely cancel out the external electric field ($E_{inside} = 0$). It acts like a perfect bridge, effectively reducing the distance $d$ to $(d-t)$.

However, when a dielectric slab is inserted, it doesn’t have free electrons. Instead, its molecules stretch (polarize), creating a weaker opposing internal field. The net electric field is reduced by a factor of $K_D$, but it isn’t zero. Hence, a dielectric increases capacitance, but not as drastically as a pure conductor of the same thickness.

$$E_{\text{net}} = E_0 – E_{\text{polarized}} = \frac{E_0}{K_D}$$

Everyday Analogy

Think of the gap $d$ as a river you have to cross. A dielectric slab is like wading through mud (thickness $t$)—it slows down the flow of the field, making it easier to store charge. A conducting slab is like building a solid bridge of length $t$—you instantly skip that part of the river, effectively shortening the gap!

Energy Density of an Electric Field

Where is the energy actually stored?

Derivation Alert

While we say a capacitor stores energy, the energy is physically stored within the electric field in the space between the plates. If we take the total energy $U = \frac{1}{2}CV^2$ and divide it by the volume between the plates ($V_{olume} = A \times d$), we get the Energy Density ($u$). This proves that any space containing an electric field contains energy.

$$u = \frac{1}{2} \varepsilon_0 E^2$$

Redistribution of Charge & Common Potential

What happens when two charged capacitors meet?

High Yield Numericals

When two charged capacitors ($C_1, V_1$ and $C_2, V_2$) are connected in parallel, charge flows from the one at higher potential to the one at lower potential until their potentials become equal. This final equalized voltage is the Common Potential ($V$). During this flow, some energy is always lost as heat in the connecting wires.

$$V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$$

$$\Delta U = \frac{C_1 C_2}{2(C_1 + C_2)} (V_1 – V_2)^2$$

Compare & Contrast

✗ Capacitors in Series

Charge ($q$) on each capacitor is the same.
Potential ($V$) divides: $V = V_1 + V_2 + V_3$.
Equivalent capacitance decreases (smaller than the smallest).
Used to decrease overall capacitance or divide high voltages.

✓ Capacitors in Parallel

Potential ($V$) across each capacitor is the same.
Charge ($q$) divides: $q = q_1 + q_2 + q_3$.
Equivalent capacitance increases (sum of all).
Used to store large amounts of charge at a low voltage.

✗ Battery Disconnected

Charge ($q$) remains constant (nowhere to go).
If dielectric is added, Potential ($V$) decreases ($V = V_0/K$).
Electric Field ($E$) decreases ($E = E_0/K$).
Stored Energy ($U$) decreases ($U = U_0/K$).

✓ Battery Remains Connected

Potential ($V$) remains constant (battery maintains it).
If dielectric is added, Charge ($q$) increases ($q = K q_0$).
Electric Field ($E$) remains constant ($E = V/d$).
Stored Energy ($U$) increases ($U = K U_0$).

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Remember

The “Battery Disconnected/Connected” table is the key to solving 90% of the reasoning questions in this chapter. Memorize what stays constant first!

Common Mistakes to Avoid

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Mistake 1

Mixing up formulas with Resistors: The formulas for capacitors in series and parallel are the exact mathematical opposites of those for resistors. Do not use $R_1+R_2$ logic for capacitors in series!

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Mistake 2

Using the wrong Energy Formula: There are three formulas: $U = \frac{1}{2}CV^2$, $U = \frac{q^2}{2C}$, and $U = \frac{1}{2}qV$.
• If $V$ is constant (battery connected), use $\frac{1}{2}CV^2$.
• If $q$ is constant (battery disconnected), use $\frac{q^2}{2C}$. Using the wrong one will lead to mathematically contradictory answers.

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Mistake 3

Ignoring units: Capacitance is almost always given in $\mu$F ($10^{-6}$ F), nF ($10^{-9}$ F), or pF ($10^{-12}$ F). You MUST convert these to standard Farads before plugging them into the energy formulas, otherwise your Joules will be off by millions.

Exam Tips

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Tip 1

When solving complex capacitor circuit diagrams, look for the “Wheatstone Bridge”. If you see 5 capacitors in a diamond shape and the ratio of arms is equal ($C_1/C_2 = C_3/C_4$), the central capacitor carries NO charge and should be erased from your diagram entirely.

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Tip 2

For the formula $C = 4 \pi \varepsilon_0 r$, memorize the value of $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9$. Therefore, $4 \pi \varepsilon_0 = \frac{1}{9 \times 10^9}$. This makes calculating the capacitance of spherical bodies like the Earth instantly solvable.

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Did You Know

The keys on older computer keyboards are actually tiny parallel plate capacitors! Pressing a key decreases the distance ($d$) between the plates. Since $C = \frac{A \varepsilon_0}{d}$, the capacitance increases, sending a digital signal to the CPU.

Expected Exam Questions

Board Pattern Questions

Class 12 · Capacitance · CBSE Exam

Class 12 · Physics

Two identical capacitors of 12 pF each are connected in series. What is the equivalent capacitance? [1 mark]

Answer 6 pF 📝

Explanation

For capacitors in series, $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$. Since they are identical, $\frac{1}{C_{eq}} = \frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6}$. Therefore, $C_{eq} = 6$ pF.
Shortcut: For ‘$n$’ identical capacitors of value $C$ in series, $C_{eq} = C/n$.

A parallel plate capacitor is charged by a battery. After charging, the battery is removed and a dielectric slab is inserted to fill the space. How will the (i) charge, (ii) potential difference, and (iii) stored energy change? [3 marks]

Answer (i) Same, (ii) Decreases, (iii) Decreases 📝

Explanation

Since the battery is disconnected, the charge $q$ is trapped and remains constant. Inserting a dielectric always increases capacitance to $C’ = KC$.
(ii) Potential difference $V’ = \frac{q}{C’} = \frac{q}{KC} = \frac{V}{K}$. It decreases by a factor of $K$.
(iii) Energy $U’ = \frac{q^2}{2C’} = \frac{q^2}{2(KC)} = \frac{U}{K}$. It decreases by a factor of $K$.

A $600$ pF capacitor is charged by a $200$ V supply. It is then disconnected from the supply and connected to another uncharged $600$ pF capacitor. How much electrostatic energy is lost in the process? [3 marks]

Answer $6 \times 10^{-6} \text{ J}$ 📝

Explanation

Given: $C_1 = 600 \text{ pF} = 600 \times 10^{-12} \text{ F}$, $V_1 = 200 \text{ V}$. $C_2 = 600 \text{ pF}$, $V_2 = 0 \text{ V}$ (uncharged).
Initial energy $U_i = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} \times 600 \times 10^{-12} \times (200)^2 = 1.2 \times 10^{-5} \text{ J}$.
When connected, common potential $V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} = \frac{600 \times 200 + 0}{600 + 600} = 100 \text{ V}$.
Final energy $U_f = \frac{1}{2} (C_1 + C_2) V^2 = \frac{1}{2} \times 1200 \times 10^{-12} \times (100)^2 = 0.6 \times 10^{-5} \text{ J}$.
Energy lost $\Delta U = U_i – U_f = 1.2 \times 10^{-5} – 0.6 \times 10^{-5} = 6 \times 10^{-6} \text{ J}$.

$$\Delta U = \frac{C_1 C_2}{2(C_1 + C_2)} (V_1 – V_2)^2 \text{ (Direct Formula)}$$

A parallel plate capacitor of capacitance $C_0$ has a conducting slab of thickness $t = d/3$ inserted between its plates. Find the new capacitance in terms of $C_0$. [2 marks]

Answer $1.5 C_0$ or $\frac{3}{2}C_0$ 📝

Explanation

The capacitance with a conducting slab is given by the formula $C = \frac{C_0}{(1 – \frac{t}{d})}$. Substituting $t = d/3$ into the equation:

$$C = \frac{C_0}{1 – \frac{d/3}{d}} = \frac{C_0}{1 – \frac{1}{3}} = \frac{C_0}{2/3} = \frac{3}{2}C_0$$

Concept Map

Capacitance connects to →

Electrostatic Potential

Parallel Plate Configuration

Dielectric Polarization

Series & Parallel Circuits

Common Potential

Electrostatic Energy

Energy Density

Capacitance – Concept Booster | Class 12 Physics CBSE

Key Concepts

Capacitance