🎯⭐ INTERACTIVE LESSON

Capacitors and Dielectrics

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Capacitors and Dielectrics - Complete Interactive Lesson

Part 1: Capacitance

⚡ Capacitance

Part 1 of 7 — Capacitance

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

Capacitance is the ratio of stored charge to voltage
SI unit: Farad (F) = C/V
Typical values: μF, nF, pF
Capacitance depends on geometry, not on $Q$ or $V$

Worked Example

A capacitor stores $6 \times 10^{-6}$ C at 3 V. Find $C$ .

$C = Q/V = 6 \times 10^{-6} / 3 = 2 \times 10^{-6}$ F μF ✅

Concept Check 🎯

Capacitance 🧮

$Q = 6$ μC, $V = 3$ V. $C$ (μF)?
$C = 4$ μF, V. (μC)?

Concept Check 🔍

Practice

#	Known	Find
1	$Q$ , $V$	$C$
2	$C$ , $V$

Challenge Question 📋

Part 2: Parallel-Plate Capacitors

⚡ Parallel-Plate Capacitors

Part 2 of 7 — Parallel-Plate Capacitors

$C = \frac{epsilon_0 A}{d}$

Part 3: Series & Parallel Combinations

⚡ Capacitors in Series & Parallel

Part 3 of 7 — Series & Parallel Combinations

Parallel: $C_{eq} = C_1 + C_2 + \cdots$ (same voltage)

Part 4: Energy Stored

⚡ Energy Stored in Capacitors

Part 4 of 7 — Energy Stored

$U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV$

Part 5: Dielectrics

⚡ Dielectrics

Part 5 of 7 — Dielectrics

Inserting a dielectric (insulating material) between plates:

$C = \kappa C_0 = \frac{\kappa epsilon_0 A}{d}$

Part 6: Problem-Solving Workshop

⚡ Problem-Solving Workshop

Part 6 of 7 — Problem-Solving Workshop

Capacitor Problem Strategy

Identify the capacitor configuration (series, parallel, or single)
Determine if a battery is connected or disconnected
Apply $Q = CV$ and energy formulas
For dielectrics, multiply $C$ by $\kappa$
Use energy density $u = \frac{1}{2} e p s i l o n_{0}$ for field energy problems

Part 7: Review & Applications

⚡ Review & Applications

Part 7 of 7 — Review & Applications

Key Formulas

$C = Q/V$ , $C = epsilon_0 A/d$ ,

$C = 2$ μF, $Q = 10$ μC. $V$ (V)?

where $A$ = plate area, $d$ = plate separation.

Electric field between plates: $E = V/d = \sigma/epsilon_0$
Field is uniform between the plates
Increasing $A$ or decreasing $d$ increases $C$

Worked Example

Plates: $A = 0.01$ m², $d = 0.001$ m. Find $C$ .

$C = epsilon_0 A/d = 8.85 \times 10^{-12} \times 0.01 / 0.001 = 88.5$ pF ✅

Concept Check 🎯

Parallel-Plate Capacitors 🧮

Plate area is doubled, $d$ unchanged. $C$ increases by factor ___
Plate area is doubled and $d$ is halved. $C$ increases by factor ___
$V = 10$ V, $d = 0.01$ m. $E$ between the plates (V/m)?

Concept Check 🔍

Practice

#	Change	Effect on $C$
1	Double $A$	$C$ doubles
2	Double $d$	$C$ halves
3	Both double	$C$ stays same

Challenge Question 📋

Series: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots$ (same charge)

Note: opposite of resistors! Parallel adds, series uses reciprocals.

Worked Example

Two capacitors: 3 μF and 6 μF in series. Find $C_{eq}$ .

$\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{1}{2}$

$C_{eq} = 2$ μF ✅

Concept Check 🎯

Series & Parallel Capacitors 🧮

3 μF and 6 μF in parallel. $C_{eq}$ (μF)?
3 μF and 6 μF in series. $C_{eq}$ (μF)?
Two 3 μF caps in parallel: $C_{eq}$ (μF)?

Concept Check 🔍

Practice

#	Configuration	Formula
1	Parallel	$C_{eq} = C_1 + C_2$
2	Series	$1/C_{eq} = 1/C_1 + 1/C_2$
3	Mixed	Simplify step by step

Challenge Question 📋

The energy is stored in the electric field between the plates.

Energy density: $u = \frac{1}{2}epsilon_0 E^2$ (J/m³)

Worked Example

A 10 μF capacitor is charged to 100 V. Energy stored?

$U = \frac{1}{2}CV^2 = \frac{1}{2}(10 \times 10^{-6})(10000) = 0.05$ J ✅

Concept Check 🎯

Energy Stored 🧮

$C = 10$ μF, $V = 100$ V. Energy (mJ)?
$V$ is doubled while $C$ stays the same. Energy increases by a factor of ___
$C = 4$ μF, $V = 10$ V. $U = \frac{1}{2}(4)(100) = ?$ μJ

Concept Check 🔍

Practice

#	Known	Formula
1	$C$ , $V$	$U = \frac{1}{2}CV^2$
2	$Q$ , $C$	$U = Q^2/(2C)$
3	$Q$ , $V$	$U = \frac{1}{2}QV$

Challenge Question 📋

where $\kappa$ (kappa) is the dielectric constant ( $\kappa > 1$ ).

Effects of a dielectric (battery disconnected):

$C$ increases by factor $\kappa$
$V$ decreases by factor $\kappa$
$E$ decreases by factor $\kappa$
$Q$ stays the same

Worked Example

A 5 μF capacitor has a dielectric with $\kappa = 3$ inserted. New capacitance?

$C = \kappa C_0 = 3 \times 5 = 15$ μF ✅

Concept Check 🎯

Dielectrics 🧮

$C_0 = 5$ μF, $\kappa = 3$ . New $C$ (μF)?
A dielectric triples the capacitance. What is $\kappa$ ?
$C_0 = 10$ μF. A dielectric with $\kappa = 2$ is inserted. Capacitor is charged to $Q = 100$ μC (battery disconnected). (V)?

Concept Check 🔍

Practice

#	Scenario	Effect
1	Insert dielectric (battery disconnected)	$C \uparrow$ , $V \downarrow$
2	Insert dielectric (battery connected)	$C \uparrow$ , $Q \uparrow$
3	Remove dielectric (battery disconnected)	$C \downarrow$ , $V \uparrow$

Challenge Question 📋

u = \frac{1}{2} e p s i l o n_{0} E^{2}

Worked Example

Three capacitors: 2 μF, 3 μF in series, then that combination in parallel with 5 μF. $V = 12$ V. Find total charge.

Series: $1/C_s = 1/2 + 1/3 = 5/6$ , $C_s = 1.2$ μF

Parallel: $C_{eq} = 1.2 + 5 = 6.2$ μF

$Q = C_{eq}V = 6.2 \times 12 = 74.4$ μC ✅

Concept Check 🎯

Problem-Solving Workshop 🧮

2 μF and 3 μF in series → 1.2 μF. In parallel with 5 μF → $C_{eq}$ (μF)? (Round to nearest integer.)
$C_{eq} = 6.2$ μF, $V = 12$ V. $Q$ (μC)? (Round to nearest integer.)
$C = 10$ μF, $V = 100$ V. Energy (mJ)?

Concept Check 🔍

Practice

#	Configuration	Key Step
1	Series-parallel	Simplify step by step
2	With dielectric	Multiply $C$ by $\kappa$
3	Energy sharing	Two capacitors connected

Challenge Question 📋

Series:

1/C_{eq} = \sum 1/C_i

Parallel:

C_{eq} = \sum C_i

Energy:

U = \frac{1}{2}CV^2 = Q^2/(2C)

Energy density:

u = \frac{1}{2}epsilon_0 E^2

Worked Example

A 20 μF capacitor with $\kappa = 5$ dielectric is charged to $V = 50$ V. Find the stored energy.

$C = \kappa C_0 = 5(20) = 100$ μF

$U = \frac{1}{2}CV^2 = \frac{1}{2}(100 \times 10^{-6})(2500) = 0.125$ J ✅

Concept Check 🎯

Review & Applications 🧮

$C = 100$ μF, $V = 50$ V. Energy (mJ)?
$\kappa = 5$ , $C_0 = 20$ μF. New $C$ (μF)?
$C_0 = 100$ pF, $\kappa = 4$ . New $C$ (pF)?

Concept Check 🔍

Practice

#	Topic	Formula
1	Capacitance	$C = Q/V$
2	Parallel plate	$C = epsilon_0 A/d$
3	Energy storage	$U = \frac{1}{2}CV^2$

Challenge Question 📋