Prerequisites
Before studying this topic, make sure you understand:
- Electric Potential — Voltage and potential difference
- Electric Field — Field between plates
- Gauss’s Law — Field due to charged surfaces
The Hook: Iron Man’s Arc Reactor
In Iron Man (2008), Tony Stark builds a miniature arc reactor that powers his entire suit — repulsor rays, flight, and all. How can such a tiny device store and deliver enormous amounts of energy?
Or think of camera flashes that charge up and release a bright burst instantly. Or defibrillators in hospitals that deliver life-saving electric shocks. Or your phone’s battery storing charge for hours of use.
The secret is capacitors — devices that store electric charge and energy, ready to release it when needed. From tiny circuits to massive power grids, capacitors are everywhere!
What is a Capacitor?
Basic principle:
- Connect capacitor to battery (voltage source)
- Positive charge accumulates on one plate
- Equal negative charge on the other plate
- Net charge = 0, but separated charge creates electric field
- Remove battery → charge stays trapped (energy stored!)
Think of a capacitor like a water tank:
| Water Tank | Capacitor |
|---|---|
| Tank size | Capacitance ($C$) |
| Water volume | Charge ($Q$) |
| Water pressure | Voltage ($V$) |
| Larger tank holds more water | Larger capacitor stores more charge |
Charging = filling the tank. Discharging = draining it.
Capacitance
where:
- $C$ = capacitance (unit: Farad, F)
- $Q$ = charge on one plate (in Coulombs)
- $V$ = potential difference between plates (in Volts)
Interactive Demo: Visualize Capacitor Charging
See how charge accumulates on the plates and how the electric field develops:
Unit: 1 Farad (F) = 1 Coulomb/Volt
Practical units:
- 1 μF = $10^{-6}$ F (microfarad)
- 1 nF = $10^{-9}$ F (nanofarad)
- 1 pF = $10^{-12}$ F (picofarad)
“Capacitance = Charge per Volt”
$C = \frac{Q}{V}$ → “C-Q-V” (like “see queue of V”)
Larger $C$ → stores more charge at same voltage
Think of it as the “capacity” of the capacitor!
Key point: Capacitance depends only on geometry (size, shape, separation) and material (dielectric), NOT on charge or voltage!
Parallel Plate Capacitor
The most common and important type.
Structure:
- Two parallel conducting plates, each of area $A$
- Separated by distance $d$
- Usually $d << \sqrt{A}$ (separation much smaller than plate dimensions)
Electric field between plates:
Using Gauss’s Law, field due to each plate with surface charge density $\sigma$:
$$E = \frac{\sigma}{\varepsilon_0}$$Potential difference:
$$V = Ed = \frac{\sigma d}{\varepsilon_0} = \frac{Qd}{A\varepsilon_0}$$(Since $\sigma = \frac{Q}{A}$)
Capacitance:
$$C = \frac{Q}{V} = \frac{Q}{Qd/(A\varepsilon_0)}$$ $$\boxed{C = \frac{\varepsilon_0 A}{d}}$$Key relationships:
- $C \propto A$ (larger area → more capacitance)
- $C \propto \frac{1}{d}$ (closer plates → more capacitance)
“Capacitance loves Close and Large”
- Closer plates ($d$ decreases) → $C$ increases
- Larger area ($A$ increases) → $C$ increases
Think of it: closer plates = stronger attraction = easier to store opposite charges!
Capacitors with Dielectric
Dielectric = insulating material (glass, plastic, ceramic) placed between plates
Effect of dielectric:
When you insert a dielectric with dielectric constant $K$ (also called relative permittivity $\varepsilon_r$):
$$\boxed{C = \frac{K\varepsilon_0 A}{d} = KC_0}$$where $C_0$ is capacitance with vacuum/air between plates.
Dielectric constant $K$:
- Vacuum: $K = 1$ (by definition)
- Air: $K \approx 1$
- Paper: $K \approx 3.7$
- Glass: $K \approx 5-10$
- Water: $K \approx 80$
- Mica: $K \approx 6$
- Ceramic: $K$ can be 1000+
Why does capacitance increase?
- Dielectric molecules get polarized in the electric field
- They create an opposing field inside
- Net field decreases: $E_{net} = \frac{E_0}{K}$
- For same charge, voltage decreases: $V = \frac{V_0}{K}$
- $C = \frac{Q}{V}$ increases by factor $K$
In Oppenheimer (2023), we see how atoms respond to electric fields — electrons shift slightly relative to the nucleus. This atomic polarization is exactly what happens in a dielectric!
Millions of atoms aligning creates a macroscopic effect — increased capacitance.
Energy Stored in a Capacitor
A charged capacitor stores electrical potential energy.
Derivation:
To charge a capacitor, we move charge $dq$ from one plate to another against potential difference $V = \frac{q}{C}$ (where $q$ is current charge).
Work done: $dW = V \, dq = \frac{q}{C} dq$
Total work (= energy stored):
$$U = \int_0^Q \frac{q}{C} dq = \frac{1}{C} \cdot \frac{Q^2}{2}$$ $$\boxed{U = \frac{Q^2}{2C}}$$Alternative forms:
Using $Q = CV$:
$$\boxed{U = \frac{1}{2}CV^2}$$ $$\boxed{U = \frac{1}{2}QV}$$All three are equivalent — use whichever is convenient!
“Half-C-V-squared, Half-Q-V, Q-squared-over-2C”
Remember the 1/2 factor — it comes from averaging (charging from 0 to V, average voltage is V/2).
Quick check: All three must give same answer!
- $\frac{1}{2}CV^2 = \frac{1}{2}(Q/V) \cdot V^2 = \frac{1}{2}QV$ ✓
- $\frac{1}{2}QV = \frac{1}{2}(CV) \cdot V = \frac{1}{2}CV^2$ ✓
Energy Density
Energy per unit volume stored in the electric field:
$$u = \frac{U}{\text{Volume}} = \frac{U}{Ad}$$Using $U = \frac{1}{2}CV^2$ and $C = \frac{\varepsilon_0 A}{d}$, $V = Ed$:
$$u = \frac{1}{2}\frac{\varepsilon_0 A}{d} \cdot \frac{(Ed)^2}{Ad} = \frac{1}{2}\varepsilon_0 E^2$$ $$\boxed{u = \frac{1}{2}\varepsilon_0 E^2}$$Key insight: Energy is stored in the electric field itself!
With dielectric:
$$\boxed{u = \frac{1}{2}K\varepsilon_0 E^2}$$Capacitors in Series
Series connection: Capacitors connected end-to-end (same charge on all).
Key properties:
- Same charge $Q$ on all capacitors
- Voltages add: $V_{total} = V_1 + V_2 + V_3 + ...$
- Equivalent capacitance:
For two capacitors:
$$\boxed{C_{eq} = \frac{C_1 C_2}{C_1 + C_2}}$$Pattern: Series capacitance is less than the smallest individual capacitance!
“Series - like Resistors in Parallel”
Series capacitors combine like resistors in parallel!
- Resistors in series: $R_{eq} = R_1 + R_2 + ...$
- Capacitors in series: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + ...$
Visual: Series capacitors = increased effective separation → less capacitance
Capacitors in Parallel
Parallel connection: Capacitors connected side-by-side (same voltage across all).
Key properties:
- Same voltage $V$ across all capacitors
- Charges add: $Q_{total} = Q_1 + Q_2 + Q_3 + ...$
- Equivalent capacitance:
Pattern: Parallel capacitance is sum of individual capacitances.
“Parallel - like Resistors in Series”
Parallel capacitors combine like resistors in series!
- Resistors in parallel: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + ...$
- Capacitors in parallel: $C_{eq} = C_1 + C_2 + ...$
Visual: Parallel capacitors = increased effective area → more capacitance
Comparison: Series vs Parallel
| Property | Series | Parallel |
|---|---|---|
| Charge | Same ($Q$) on all | Different (adds up) |
| Voltage | Adds up | Same ($V$) across all |
| Equivalent $C$ | $\frac{1}{C_{eq}} = \sum \frac{1}{C_i}$ | $C_{eq} = \sum C_i$ |
| Comparison | $C_{eq} <$ smallest $C_i$ | $C_{eq} >$ largest $C_i$ |
| Energy | $U = \frac{Q^2}{2C_{eq}}$ | $U = \frac{1}{2}C_{eq}V^2$ |
Think of Jawan (2023) team planning:
Series: Each person handles part of the task sequentially → total time increases (like voltage adds up)
Parallel: Everyone works together → combined capacity increases (like capacitance adds up)
Same logic applies to capacitors!
Effects of Dielectric: Two Cases
Case 1: Battery Connected (Constant Voltage)
When you insert dielectric while battery is connected:
Before: $Q_0 = C_0 V$, $U_0 = \frac{1}{2}C_0 V^2$
After:
- Voltage $V$ stays same (battery maintains it)
- Capacitance: $C = KC_0$
- Charge: $Q = CV = KC_0 V$ (increases by factor $K$)
- Energy: $U = \frac{1}{2}CV^2 = K \cdot \frac{1}{2}C_0 V^2$ (increases by factor $K$)
Summary: $Q$ and $U$ increase by factor $K$
Case 2: Battery Disconnected (Constant Charge)
When you insert dielectric after disconnecting battery:
Before: $Q_0 = C_0 V_0$, $U_0 = \frac{Q_0^2}{2C_0}$
After:
- Charge $Q = Q_0$ stays same (isolated capacitor)
- Capacitance: $C = KC_0$
- Voltage: $V = \frac{Q}{C} = \frac{Q_0}{KC_0} = \frac{V_0}{K}$ (decreases by factor $K$)
- Energy: $U = \frac{Q^2}{2C} = \frac{Q_0^2}{2KC_0} = \frac{U_0}{K}$ (decreases by factor $K$)
Summary: $V$ and $U$ decrease by factor $K$
Always check: Is battery connected or disconnected?
- Connected → $V$ constant, $Q$ changes
- Disconnected → $Q$ constant, $V$ changes
This determines whether energy increases or decreases!
Common Capacitor Configurations
1. Spherical Capacitor
Two concentric spheres, radii $a$ (inner) and $b$ (outer):
$$C = 4\pi\varepsilon_0 \frac{ab}{b-a}$$For $b >> a$: $C \approx 4\pi\varepsilon_0 a$ (isolated sphere)
2. Cylindrical Capacitor
Two coaxial cylinders, radii $a$ (inner) and $b$ (outer), length $L$:
$$C = \frac{2\pi\varepsilon_0 L}{\ln(b/a)}$$3. Parallel Plates (main formula)
$$C = \frac{\varepsilon_0 A}{d}$$Charging and Discharging (RC Circuits)
When capacitor charges through resistor $R$ from voltage source $V_0$:
Charging:
- Charge: $Q(t) = CV_0(1 - e^{-t/RC})$
- Voltage: $V_C(t) = V_0(1 - e^{-t/RC})$
- Current: $I(t) = \frac{V_0}{R}e^{-t/RC}$
Discharging:
- Charge: $Q(t) = Q_0 e^{-t/RC}$
- Voltage: $V_C(t) = V_0 e^{-t/RC}$
- Current: $I(t) = -\frac{V_0}{R}e^{-t/RC}$
Time constant: $\tau = RC$
- At $t = \tau$: capacitor charges to 63% of maximum
- At $t = 5\tau$: essentially fully charged (>99%)
Common Mistakes to Avoid
Wrong: Series capacitors add: $C_{series} = C_1 + C_2$
Correct:
- Parallel: $C = C_1 + C_2$ (adds directly)
- Series: $\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$ (reciprocals add)
Opposite of resistors!
Wrong: Using wrong formula for given quantities
Correct: Pick the right one:
- Given $Q$ and $C$: Use $U = \frac{Q^2}{2C}$
- Given $V$ and $C$: Use $U = \frac{1}{2}CV^2$
- Given $Q$ and $V$: Use $U = \frac{1}{2}QV$
Wrong: “Dielectric always increases energy”
Correct:
- Battery connected: Energy increases ($U = \frac{1}{2}CV^2$, $C$ increases)
- Battery disconnected: Energy decreases ($U = \frac{Q^2}{2C}$, $C$ increases)
Check if $Q$ or $V$ is constant!
Wrong: $C = \frac{\varepsilon_0 d}{A}$
Correct: $C = \frac{\varepsilon_0 A}{d}$
Area in numerator, distance in denominator!
Practice Problems
Level 1: Foundation (NCERT)
A parallel plate capacitor has plates of area $100$ cm² separated by 2 mm. Find its capacitance.
Solution:
$$C = \frac{\varepsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 100 \times 10^{-4}}{2 \times 10^{-3}}$$ $$C = \frac{8.85 \times 10^{-12} \times 10^{-2}}{2 \times 10^{-3}} = \frac{8.85 \times 10^{-14}}{2 \times 10^{-3}}$$ $$C = 4.425 \times 10^{-11} \text{ F}$$ $$\boxed{C \approx 44.25 \text{ pF}}$$A 10 μF capacitor is charged to 100 V. Find (a) charge stored, (b) energy stored.
Solution:
(a) Charge:
$$Q = CV = 10 \times 10^{-6} \times 100 = 10^{-3} \text{ C}$$ $$\boxed{Q = 1 \text{ mC}}$$(b) Energy:
$$U = \frac{1}{2}CV^2 = \frac{1}{2} \times 10 \times 10^{-6} \times (100)^2$$ $$U = 5 \times 10^{-6} \times 10^4 = 5 \times 10^{-2} \text{ J}$$ $$\boxed{U = 0.05 \text{ J} = 50 \text{ mJ}}$$Two capacitors of 4 μF and 6 μF are connected in series. Find equivalent capacitance.
Solution:
$$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{4} + \frac{1}{6} = \frac{3 + 2}{12} = \frac{5}{12}$$ $$C_{eq} = \frac{12}{5} = 2.4 \text{ μF}$$ $$\boxed{C_{eq} = 2.4 \text{ μF}}$$(Less than smallest: 2.4 < 4 ✓)
Level 2: JEE Main
A parallel plate capacitor is charged to 200 V and then disconnected. A dielectric slab ($K = 5$) is inserted. Find the new voltage.
Solution:
Battery disconnected → Charge $Q$ constant
Original: $V_0 = 200$ V
After inserting dielectric:
- Capacitance becomes $C = KC_0 = 5C_0$
- Charge same: $Q = C_0 V_0 = CV$
(Voltage decreases by factor 5)
Three capacitors 2 μF, 3 μF, and 6 μF are connected in parallel. This combination is connected in series with a 4 μF capacitor. Find total capacitance.
Solution:
Step 1: Parallel combination:
$$C_{parallel} = 2 + 3 + 6 = 11 \text{ μF}$$Step 2: Series with 4 μF:
$$\frac{1}{C_{total}} = \frac{1}{11} + \frac{1}{4} = \frac{4 + 11}{44} = \frac{15}{44}$$ $$C_{total} = \frac{44}{15} \approx 2.93 \text{ μF}$$ $$\boxed{C_{total} \approx 2.93 \text{ μF}}$$A capacitor of capacitance $C$ is charged to voltage $V$ and then connected to an identical uncharged capacitor. Find (a) final voltage on each, (b) energy loss.
Solution:
Initial charge: $Q_0 = CV$
When connected, charge distributes equally (same voltage on both):
(a) Final voltage:
Total charge conserved: $Q_0 = Q_1 + Q_2 = CV_f + CV_f = 2CV_f$
$$V_f = \frac{Q_0}{2C} = \frac{CV}{2C} = \frac{V}{2}$$ $$\boxed{V_f = \frac{V}{2} \text{ on each}}$$(b) Energy loss:
Initial energy: $U_i = \frac{1}{2}CV^2$
Final energy: $U_f = 2 \times \frac{1}{2}C\left(\frac{V}{2}\right)^2 = 2 \times \frac{1}{2}C \times \frac{V^2}{4} = \frac{CV^2}{4}$
Energy loss: $\Delta U = U_i - U_f = \frac{1}{2}CV^2 - \frac{1}{4}CV^2$
$$\boxed{\Delta U = \frac{1}{4}CV^2}$$(Half the energy is lost as heat during charge redistribution!)
Level 3: JEE Advanced
A capacitor is made of two square plates of side $L$ separated by distance $d$. A dielectric slab of thickness $d$ and dielectric constant $K$ is inserted to fill half the gap (covering half the area). Find capacitance.
Solution:
Think of it as two capacitors in parallel:
Capacitor 1 (with dielectric, area $A/2 = L^2/2$):
$$C_1 = \frac{K\varepsilon_0 (L^2/2)}{d} = \frac{K\varepsilon_0 L^2}{2d}$$Capacitor 2 (without dielectric, area $A/2 = L^2/2$):
$$C_2 = \frac{\varepsilon_0 (L^2/2)}{d} = \frac{\varepsilon_0 L^2}{2d}$$Total (parallel):
$$C = C_1 + C_2 = \frac{K\varepsilon_0 L^2}{2d} + \frac{\varepsilon_0 L^2}{2d}$$ $$C = \frac{\varepsilon_0 L^2}{2d}(K + 1)$$ $$\boxed{C = \frac{\varepsilon_0 L^2(K+1)}{2d}}$$An infinite number of capacitors are connected as shown: first $C$, then $2C$ in series with it, then $4C$ in series, then $8C$, etc. Find effective capacitance.
Solution:
Let effective capacitance be $C_{eq}$.
After the first capacitor $C$, the rest of the network (starting from $2C$) is similar to the original but with capacitances doubled.
So, rest of network has capacitance $2C_{eq}$ (by scaling).
First capacitor $C$ is in series with $2C_{eq}$:
$$\frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{2C_{eq}}$$ $$\frac{1}{C_{eq}} - \frac{1}{2C_{eq}} = \frac{1}{C}$$ $$\frac{2 - 1}{2C_{eq}} = \frac{1}{C}$$ $$\frac{1}{2C_{eq}} = \frac{1}{C}$$ $$\boxed{C_{eq} = 2C}$$Beautiful result: Infinite series converges to finite value!
A parallel plate capacitor with air ($K=1$) has capacitance $C_0$ and is charged to voltage $V_0$, then disconnected. A dielectric slab ($K=4$) of thickness $d/2$ (where $d$ is plate separation) is inserted. Find (a) new capacitance, (b) new voltage, (c) new energy.
Solution:
Think as two capacitors in series:
- Gap with dielectric (thickness $d/2$): $C_1 = \frac{K\varepsilon_0 A}{d/2} = \frac{4 \times 2\varepsilon_0 A}{d} = \frac{8\varepsilon_0 A}{d}$
- Gap with air (thickness $d/2$): $C_2 = \frac{\varepsilon_0 A}{d/2} = \frac{2\varepsilon_0 A}{d}$
(a) New capacitance:
$$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d}{8\varepsilon_0 A} + \frac{d}{2\varepsilon_0 A} = \frac{d}{\varepsilon_0 A}\left[\frac{1}{8} + \frac{4}{8}\right]$$ $$\frac{1}{C} = \frac{5d}{8\varepsilon_0 A}$$ $$C = \frac{8\varepsilon_0 A}{5d}$$Since $C_0 = \frac{\varepsilon_0 A}{d}$:
$$\boxed{C = \frac{8C_0}{5} = 1.6C_0}$$(b) New voltage:
Charge conserved: $Q = C_0 V_0$
$$V = \frac{Q}{C} = \frac{C_0 V_0}{1.6C_0} = \frac{V_0}{1.6}$$ $$\boxed{V = \frac{5V_0}{8} = 0.625V_0}$$(c) New energy:
$$U = \frac{Q^2}{2C} = \frac{(C_0 V_0)^2}{2 \times 1.6C_0} = \frac{C_0 V_0^2}{3.2}$$Original: $U_0 = \frac{1}{2}C_0 V_0^2$
$$U = \frac{U_0}{1.6} = \frac{5U_0}{8}$$ $$\boxed{U = 0.625U_0}$$(Energy decreases by 37.5%)
Quick Revision Box
| Concept | Formula/Key Point |
|---|---|
| Capacitance | $C = \frac{Q}{V}$ (charge per volt) |
| Parallel plate | $C = \frac{\varepsilon_0 A}{d}$ (or $\frac{K\varepsilon_0 A}{d}$ with dielectric) |
| Energy stored | $U = \frac{1}{2}CV^2 = \frac{1}{2}QV = \frac{Q^2}{2C}$ |
| Energy density | $u = \frac{1}{2}\varepsilon_0 E^2$ |
| Series combination | $\frac{1}{C_{eq}} = \sum \frac{1}{C_i}$ (like R in parallel) |
| Parallel combination | $C_{eq} = \sum C_i$ (like R in series) |
| Two in series | $C = \frac{C_1 C_2}{C_1 + C_2}$ |
| Dielectric effect | $C$ multiplied by $K$ |
| Battery connected | $V$ constant, $Q$ and $U$ increase with dielectric |
| Battery disconnected | $Q$ constant, $V$ and $U$ decrease with dielectric |
| Time constant | $\tau = RC$ (charging/discharging) |
JEE Weightage: 2-3 questions in JEE Main, 1-2 in JEE Advanced (often numerical on combinations or energy)
Time-Saver: For series/parallel, remember it’s opposite of resistors!
Teacher’s Summary
- Capacitance is geometry-dependent — larger area and closer plates give more capacitance
- Capacitors store energy in electric field — $U = \frac{1}{2}CV^2$
- Series and parallel are opposite of resistors — series adds reciprocals, parallel adds directly
- Dielectric increases capacitance — factor $K$ (dielectric constant)
- Battery connected vs disconnected matters — determines if $V$ or $Q$ is constant
“Capacitors are the energy reservoirs of electronics. Master their combinations and you master circuit design.”
Related Topics
Within Electrostatics
- Electric Potential — Voltage across capacitor
- Electric Field — Field between capacitor plates
- Gauss’s Law — Deriving field for parallel plates
Connected Chapters
- Current Electricity — RC circuits, charging/discharging
- Electromagnetic Waves — Oscillating LC circuits
- AC Circuits — Capacitive reactance
Applications
- Camera flash storage
- Computer memory (DRAM uses capacitors)
- Power supply filters
- Touch screens (capacitive sensing)
- Defibrillators (energy storage)
Math Connections
- Exponential Functions — RC circuit equations
- Differential Equations — Charging/discharging dynamics
Final Note
Congratulations! You’ve completed the Electrostatics chapter. You now understand:
- How charges interact (Coulomb’s Law)
- How they create fields and potentials
- How dipoles behave
- How to use Gauss’s Law for symmetric problems
- How to work with energy and potential
- How capacitors store charge and energy
Next steps:
- Practice more problems from previous JEE papers
- Connect these concepts to Current Electricity
- Explore applications in Magnetism and Electromagnetic Induction
Keep practicing, and remember: Electrostatics is the foundation for all of electromagnetism!