Electrostatics

LessonAbout 1.5 h

Electrostatics is the branch of physics that studies electric charges at rest and their surrounding fields.

1. Coulomb’s Law

Coulomb’s law states that the electrostatic force between two point charges is directly proportional to the product of charges and inversely proportional to the square of distance between them.

First law: like charges repel each other, unlike charges attract each other
Second law: force varies directly with charge product and inversely with squared distance
Dielectric constant: relative permittivity \(\epsilon_r\) reduces electrostatic attraction force

Coulomb’s Law

\[ F = \dfrac{Q_1 Q_2}{4 \pi \epsilon_0 \epsilon_r d^2} = \dfrac{k Q_1 Q_2}{d^2} \]

Where: \(k = 9 \times 10^9\ \text{N}\cdot\text{m}^2/\text{C}^2\), \(\epsilon_0 = 8.854 \times 10^{-12}\ \text{F/m}\)

EE Board Exam April 1997

Example. Two electrons in a vacuum experience a force of \(2 \times 10^{-15}\ \text{N}\). How far apart are the electrons?

Solution.

\[ d = \sqrt{\dfrac{k Q^2}{F}} = \sqrt{\dfrac{(9 \times 10^9)(1.6 \times 10^{-19}\ \text{C})^2}{2 \times 10^{-15}\ \text{N}}} = 3.39 \times 10^{-7}\ \text{m} \]

Answer: \(3.39 \times 10^{-7}\ \text{m}\)

2. Electrostatic Potential

Electrostatic potential is the work done in bringing a unit positive charge from infinity to a point in an electric field.

Unit: volt (\(\text{V}\)) or joule per coulomb (\(\text{J/C}\))
Scalar quantity: cumulative potential from multiple charges sums algebraically
Reference level: zero potential assumed at infinite distance

Electrostatic Potential

\[ V = \dfrac{Q}{4 \pi \epsilon_0 \epsilon_r d} \]

Example. A battery can deliver \(10\ \text{J}\) of energy to move \(5\ \text{C}\) of charge. What is the potential difference between the terminals of the battery?

Solution.

\[ V = \dfrac{W}{Q} = \dfrac{10\ \text{J}}{5\ \text{C}} = 2\ \text{V} \]

Answer: \(2\ \text{V}\)

3. Electric Field Intensity

Electric field intensity is the force per unit positive test charge placed at a point in an electric field.

Unit: newton per coulomb (\(\text{N/C}\)) or volt per meter (\(\text{V/m}\))
Vector quantity: directed outward from positive charge and inward toward negative charge
Field of isolated conductor: concentrated uniformly over conducting outer boundary

Electric Field Intensity

\[ E = \dfrac{Q}{4 \pi \epsilon_0 \epsilon_r d^2} \]

Field Outside Isolated Sphere

\[ E = \dfrac{Q}{4 \pi \epsilon_0 r^2} \]

Field of Isolated Cylindrical Wire

\[ E = \dfrac{\Delta}{2 \pi \epsilon_0 r} \]

Where: \(\Delta\) = charge per unit length (\(\text{C/m}\))

EE Board Exam October 1998

Example. Calculate the magnitude of the electric field at a point that is \(30\ \text{cm}\) from a point charge \(Q = -3.2 \times 10^{-6}\ \text{C}\).

Solution.

\[ E = \dfrac{k Q}{d^2} = \dfrac{(9 \times 10^9)(3.2 \times 10^{-6}\ \text{C})}{(0.3\ \text{m})^2} = 3.2 \times 10^5\ \text{N/C} \]

Answer: \(3.2 \times 10^5\ \text{N/C}\)

4. Electric Flux Density

Electric flux density is the electric flux passing perpendicularly through a unit surface area.

Unit: coulomb per square meter (\(\text{C/m}^2\))
Gauss’s relation: total electric flux \(\Psi\) exiting closed surface equals enclosed charge \(Q\)
Medium independence: electric flux density \(D\) depends on charge distribution, independent of medium

Electric Flux Density

\[ D = \dfrac{\Psi}{A} = \epsilon_0 \epsilon_r E \]

Where: \(\Psi\) = electric flux (\(\text{C}\))

5. Capacitance

Capacitance is the ability of a conductor system or capacitor to store electric charge per unit potential difference.

Unit: farad (\(\text{F}\)) or coulomb per volt (\(\text{C/V}\))
Dielectric material: insulating layer separating conducting plates
Charge storage: equal and opposite charges accumulated on opposing plates

Capacitance

\[ C = \dfrac{Q}{V} \]

EE Board Exam October 1990

Example. A capacitor whose plates are \(20\ \text{cm} \times 3.0\ \text{cm}\) and separated by a \(1.0\text{-}\text{mm}\) air gap is connected across a \(12\text{-}\text{V}\) battery. Determine the charge accumulated on each plate after a long time.

Solution.

\[ C = \dfrac{\epsilon_0 A}{d} = \dfrac{(8.854 \times 10^{-12})(0.20 \times 0.03)}{0.001} = 53.124 \times 10^{-12}\ \text{F} \]

\[ Q = C V = (53.124 \times 10^{-12}\ \text{F})(12\ \text{V}) = 6.4 \times 10^{-10}\ \text{C} \]

Answer: \(6.4 \times 10^{-10}\ \text{C}\)

5.1 Two Parallel Plates

Two parallel-plate capacitor consists of two conducting plates of area \(A\) separated by distance \(d\) in air or vacuum.

Two Parallel-Plate Capacitor

\[ C = \dfrac{\epsilon_0 A}{d} \]

EE Board Exam October 1991

Example. Calculate the capacitance between two parallel plates each of which is \(100\ \text{cm}^2\) and \(2\ \text{mm}\) apart in air.

Solution.

\[ C = \dfrac{\epsilon_0 A}{d} = \dfrac{(8.854 \times 10^{-12})(100 \times 10^{-4}\ \text{m}^2)}{0.002\ \text{m}} = 0.443 \times 10^{-12}\ \text{F} = 0.443\ \mu\mu\text{F} \]

Answer: \(0.443\ \mu\mu\text{F}\)

5.2 N Parallel Plates

Multi-plate capacitor consists of \(n\) parallel interleaved conducting plates forming \((n - 1)\) parallel dielectric spaces.

Multi-Plate Capacitor

\[ C = \dfrac{\epsilon_0 \epsilon_r A}{d} (n - 1) \]

EE Board Exam October 1994

Example. A parallel plate capacitor is made of \(350\) plates, separated by paraffined paper \(0.0010\ \text{cm}\) thick (\(k = 2.5\)). The effective size of the plate is \(15 \times 30\ \text{cm}\). What is the capacitance of this capacitor?

Solution.

\[ C = \dfrac{\epsilon_0 \epsilon_r A}{d} (n - 1) = \dfrac{(8.854 \times 10^{-12})(2.5)(0.15 \times 0.30)}{0.00001} (350 - 1) = 34.76\ \mu\text{F} \approx 35\ \mu\text{F} \]

Answer: \(35\ \mu\text{F}\)

5.3 Parallel Plate with Medium

Capacitor with composite dielectric contains multiple insulating layers or partial air gap between parallel plates.

Capacitance with Multiple Dielectrics

\[ C = \dfrac{\epsilon_0 A}{\dfrac{d_1}{\epsilon_{r1}} + \dfrac{d_2}{\epsilon_{r2}} + \dfrac{d_3}{\epsilon_{r3}}} \]

Capacitor with Partially Air Dielectric

\[ C = \dfrac{\epsilon_0 A}{d - t \left(1 - \dfrac{1}{\epsilon_r}\right)} \]

Where: \(t\) = thickness of dielectric slab inserted into total gap \(d\)

Example. A capacitor has two parallel metal plates measuring \(0.45\ \text{m}\) square each. The distance between the plates is \(1\ \text{cm}\) in free air. If a dielectric material whose permittivity is \(4\) and \(5\ \text{mm}\) thick is placed in the upper plate, leaving an air gap, what is the capacitance?

Solution.

\[ C = \dfrac{\epsilon_0 A}{d - t \left(1 - \dfrac{1}{\epsilon_r}\right)} = \dfrac{(8.854 \times 10^{-12})(0.45 \times 0.45)}{0.01 - 0.005 \left(1 - \dfrac{1}{4}\right)} = 287\ \mu\mu\text{F} \]

Answer: \(287\ \mu\mu\text{F}\)

5.4 Isolated Sphere

Isolated spherical conductor holds charge in free space with infinite surrounding reference plane.

Capacitance of Isolated Sphere

\[ C = 4 \pi \epsilon_0 \epsilon_r r \]

5.5 Concentric Spheres

Concentric spherical capacitor consists of two concentric conducting spherical shells separated by dielectric material.

Concentric Spherical Capacitor

\[ C = 4 \pi \epsilon_0 \epsilon_r \dfrac{r_1 r_2}{r_2 - r_1} \]

5.6 Coaxial Cable

Coaxial cable capacitance represents capacitance per unit length between inner cylindrical conductor and outer metallic sheath.

Coaxial Cable Capacitance

\[ C = \dfrac{2 \pi \epsilon_0 \epsilon_r}{\ln\left(\dfrac{r_2}{r_1}\right)} \]

EE Board Exam April 1997

Example. A coaxial cable has a \(10\text{-}\text{mm}\) diameter inside conductor and a metallic sheath with an inside diameter of \(20\ \text{mm}\). If the insulating medium has a dielectric constant \(k = 2\), calculate the capacitance between the conductor and shield per km.

Solution.

\[ C = \dfrac{2 \pi \epsilon_0 \epsilon_r}{\ln\left(\dfrac{r_2}{r_1}\right)} = \dfrac{2 \pi (8.854 \times 10^{-12})(2)}{\ln\left(\dfrac{10}{5}\right)} = 1.605 \times 10^{-10}\ \text{F/m} = 0.1605\ \mu\text{F/km} \]

Answer: \(0.1605\ \mu\text{F/km}\)

5.7 Two-Wire Line with Opposite Charges

Parallel-wire line capacitance measures line-to-line capacitance per unit length between two parallel cylindrical conductors carrying equal and opposite charges.

Two-Wire Line Capacitance

\[ C = \dfrac{\pi \epsilon_0 \epsilon_r}{\ln\left(\dfrac{D}{r}\right)} \]

6. Series Capacitors

Series capacitors are connected end-to-end across a source, carrying equal accumulated electric charge.

Charge equality: identical charge accumulates on every series capacitor (\(Q_T = Q_1 = Q_2\))
Voltage summation: total applied voltage equals sum of series voltage drops
Equivalent reciprocal: inverse total capacitance equals sum of inverse individual capacitances

Series Capacitance

\[ \dfrac{1}{C_T} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dots + \dfrac{1}{C_n} \]

\[ Q_T = Q_1 = Q_2 = \dots = Q_n \]

EE Board Exam October 1990

Example. The result of capacitance \(C_1 = 6\ \mu\text{F}\) and \(C_2\) connected in series is \(3\ \mu\text{F}\). Find capacitor \(C_2\) in microfarads.

Solution.

\[ \dfrac{1}{C_T} = \dfrac{1}{C_1} + \dfrac{1}{C_2} \implies \dfrac{1}{3} = \dfrac{1}{6} + \dfrac{1}{C_2} \implies C_2 = 6\ \mu\text{F} \]

Answer: \(6\ \mu\text{F}\)

7. Voltage Division

Voltage division theorem for capacitors distributes total voltage inversely proportional to individual capacitances in a series circuit.

Inverse proportion: smaller capacitance receives higher voltage drop
Two-capacitor rule: voltage drop across \(C_1\) depends directly on opposite capacitance \(C_2\)
Charge conservation: common charge equals product of equivalent capacitance and total voltage

Voltage Division for Capacitors

\[ V_1 = V_T \dfrac{C_2}{C_1 + C_2} \qquad\qquad V_2 = V_T \dfrac{C_1}{C_1 + C_2} \]

Example. Three capacitors of \(20\), \(10\) and \(15\ \mu\text{F}\) respectively are connected in series across a \(100\text{-}\text{V}\) supply. Find the voltage across the \(10\ \mu\text{F}\) capacitor.

Solution.

\[ \dfrac{1}{C_T} = \dfrac{1}{20} + \dfrac{1}{10} + \dfrac{1}{15} = 0.21666\ \mu\text{F}^{-1} \implies C_T = 4.615\ \mu\text{F} \]

\[ Q_T = C_T V_T = (4.615\ \mu\text{F})(100\ \text{V}) = 461.5\ \mu\text{C} \]

\[ V_2 = \dfrac{Q_T}{C_2} = \dfrac{461.5\ \mu\text{C}}{10\ \mu\text{F}} = 46.15\ \text{V} \]

Answer: \(46.15\ \text{V}\)

8. Parallel Capacitors

Parallel capacitors are connected across common nodes, sharing identical applied terminal voltage.

Voltage equality: identical voltage exists across all parallel branches (\(V_T = V_1 = V_2\))
Charge summation: total charge equals sum of individual branch charges
Direct summation: total equivalent capacitance equals sum of parallel capacitances

Parallel Capacitance

\[ C_T = C_1 + C_2 + \dots + C_n \]

\[ V_T = V_1 = V_2 = \dots = V_n \]

Example. Four capacitors A, B, C and D are connected across a \(150\text{-}\text{V}\) DC source. The charges accumulated in A, B and C are equal to \(2000\), \(750\) and \(1500\ \mu\text{C}\), respectively. If the total capacitance of the combination is \(35\ \mu\text{F}\), how much charge is accumulated in capacitor D?

Solution.

\[ Q_T = C_T V_T = (35\ \mu\text{F})(150\ \text{V}) = 5250\ \mu\text{C} \]

\[ Q_D = Q_T - (Q_A + Q_B + Q_C) = 5250 - (2000 + 750 + 1500) = 1000\ \mu\text{C} \]

Answer: \(1000\ \mu\text{C}\)

9. Charge Division

Charge division theorem for capacitors divides total parallel charge directly proportional to individual capacitances.

Direct proportion: larger capacitance stores larger fraction of total charge
Charge redistribution: rejoining charged capacitors causes charge transfer until voltages equalize
Polarity rule: like polarities connected together sum initial charges algebraically

Charge Division for Capacitors

\[ Q_1 = Q_T \dfrac{C_1}{C_1 + C_2} \qquad\qquad Q_2 = Q_T \dfrac{C_2}{C_1 + C_2} \]

Example. Two capacitors with capacitance of \(6\ \mu\text{F}\) and \(4\ \mu\text{F}\) are connected in series across a \(100\text{-}\text{V}\) DC source. If the supply voltage is cut off and the two capacitors are connected in parallel, what will be the final charge accumulated in the \(4\ \mu\text{F}\) capacitor?

Solution.

\[ C_s = \dfrac{6 \times 4}{6 + 4} = 2.4\ \mu\text{F} \qquad\qquad Q_T = C_s V = (2.4)(100) = 240\ \mu\text{C} \qquad\qquad Q_{\text{total}} = 240 + 240 = 480\ \mu\text{C} \]

\[ Q_2 = Q_{\text{total}} \dfrac{C_2}{C_1 + C_2} = 480 \left(\dfrac{4}{6 + 4}\right) = 192\ \mu\text{C} \]

Answer: \(192\ \mu\text{C}\)

10. Energy Stored

Electrostatic energy stored in a capacitor represents potential energy stored in the dielectric electric field.

Work done: charging process transfers work into potential field energy
Energy loss: reconnecting unequal voltage capacitors dissipates spark/heat energy
Voltage dependence: stored energy varies with square of applied terminal voltage

Energy Stored in Capacitor

\[ W = \dfrac{1}{2} C V^2 = \dfrac{1}{2} Q V = \dfrac{1}{2} \dfrac{Q^2}{C} \]

EE Board Exam October 1997

Example. A \(20\ \mu\text{F}\) capacitor is charged by a \(12\text{-}\text{V}\) battery. What is the stored energy in the capacitor?

Solution.

\[ W = \dfrac{1}{2} C V^2 = \dfrac{1}{2} (20 \times 10^{-6}\ \text{F})(12\ \text{V})^2 = 1.44 \times 10^{-3}\ \text{J} \]

Answer: \(1.44 \times 10^{-3}\ \text{J}\)

11. Elastance

Elastance is the reciprocal of capacitance, measuring opposition to storing electric charge.

Unit: daraf (\(\text{daraf}\)) or reciprocal farad (\(\text{F}^{-1}\))
Series addition: total elastance of series capacitors equals sum of individual elastances
Relation: elastance increases as dielectric thickness between plates increases

Elastance

\[ S = \dfrac{1}{C} \]

Where: \(S\) = elastance (\(\text{daraf}\) or \(\text{F}^{-1}\))

Formula Summary

Formula Name	Formula
Coulomb’s Law	\(F = \dfrac{Q_1 Q_2}{4 \pi \epsilon_0 \epsilon_r d^2} = \dfrac{k Q_1 Q_2}{d^2}\)
Electrostatic Potential	\(V = \dfrac{Q}{4 \pi \epsilon_0 \epsilon_r d}\)
Electric Field Intensity	\(E = \dfrac{Q}{4 \pi \epsilon_0 \epsilon_r d^2}\)
Field Outside Isolated Sphere	\(E = \dfrac{Q}{4 \pi \epsilon_0 r^2}\)
Field of Isolated Cylindrical Wire	\(E = \dfrac{\Delta}{2 \pi \epsilon_0 r}\)
Electric Flux Density	\(D = \dfrac{\Psi}{A} = \epsilon_0 \epsilon_r E\)
Capacitance	\(C = \dfrac{Q}{V}\)
Two Parallel-Plate Capacitor	\(C = \dfrac{\epsilon_0 A}{d}\)
Multi-Plate Capacitor	\(C = \dfrac{\epsilon_0 \epsilon_r A}{d} (n - 1)\)
Capacitance with Multiple Dielectrics	\(C = \dfrac{\epsilon_0 A}{\dfrac{d_1}{\epsilon_{r1}} + \dfrac{d_2}{\epsilon_{r2}} + \dfrac{d_3}{\epsilon_{r3}}}\)
Capacitor with Partially Air Dielectric	\(C = \dfrac{\epsilon_0 A}{d - t \left(1 - \dfrac{1}{\epsilon_r}\right)}\)
Capacitance of Isolated Sphere	\(C = 4 \pi \epsilon_0 \epsilon_r r\)
Concentric Spherical Capacitor	\(C = 4 \pi \epsilon_0 \epsilon_r \dfrac{r_1 r_2}{r_2 - r_1}\)
Coaxial Cable Capacitance	\(C = \dfrac{2 \pi \epsilon_0 \epsilon_r}{\ln\left(\dfrac{r_2}{r_1}\right)}\)
Two-Wire Line Capacitance	\(C = \dfrac{\pi \epsilon_0 \epsilon_r}{\ln\left(\dfrac{D}{r}\right)}\)
Series Capacitance	\(\dfrac{1}{C_T} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dots + \dfrac{1}{C_n}\)
Voltage Division for Capacitors	\(V_1 = V_T \dfrac{C_2}{C_1 + C_2} \quad V_2 = V_T \dfrac{C_1}{C_1 + C_2}\)
Parallel Capacitance	\(C_T = C_1 + C_2 + \dots + C_n\)
Charge Division for Capacitors	\(Q_1 = Q_T \dfrac{C_1}{C_1 + C_2} \quad Q_2 = Q_T \dfrac{C_2}{C_1 + C_2}\)
Energy Stored in Capacitor	\(W = \dfrac{1}{2} C V^2 = \dfrac{1}{2} Q V = \dfrac{1}{2} \dfrac{Q^2}{C}\)
Elastance	\(S = \dfrac{1}{C}\)