Electric Charges and Fields - Comprehensive Notes

Introduction to Electric Charges and Fields

• Experiencing sparks or crackling sounds when removing synthetic clothes, especially in dry weather, is due to electric discharge.
• Lightning is a common example of electric discharge.
• Electric shocks from car doors or bus bars are due to accumulated electric charges discharging through the body.
• These phenomena are related to static electricity, which is the focus of this chapter.
• Electrostatics studies forces, fields, and potentials arising from static charges.

Electric Charge

• Thales of Miletus (600 BC) discovered that amber rubbed with wool attracts light objects.
• The word "electricity" comes from the Greek word "elektron," meaning amber.
• Rubbing certain pairs of materials can attract light objects like straw and paper.
• Glass rods rubbed with silk repel each other, and plastic rods rubbed with fur also repel each other.
• However, a glass rod and fur attract each other; plastic rod and glass rod attract each other.
• Careful experiments showed only two kinds of electric charge exist.
• Objects like glass or plastic rods become electrified when rubbed, acquiring an electric charge.
• Like charges repel, and unlike charges attract each other.
• Polarity of charge differentiates the two kinds of charges.
• When a glass rod is rubbed with silk, they acquire opposite charges.
• Bringing an electrified glass rod into contact with silk neutralizes their charges.
• Benjamin Franklin named the charges positive and negative.
• By convention, glass rod or cat's fur is positive, and plastic rod or silk is negative.
• An object with an electric charge is electrified or charged; without charge, it is electrically neutral.

Detecting Charge

• A gold-leaf electroscope detects charge.
• It consists of a vertical metal rod in a box with two thin gold leaves at the bottom.
• When a charged object touches the metal knob, charge flows to the leaves, causing them to diverge.
• The degree of divergence indicates the amount of charge.

Understanding How Materials Acquire Charge

• Matter is made of atoms and molecules containing charges that are normally balanced.
• Forces holding molecules together, adhesive forces, and surface tension forces are electrical.
• Electric force is pervasive in life.
• Electrifying a neutral body involves adding or removing one kind of charge, referring to excess or deficit of charge.
• In solids, loosely bound electrons transfer between bodies.
• A body can be positively charged by losing electrons or negatively charged by gaining electrons.
• When rubbing a glass rod with silk, electrons transfer from the rod to the silk.
• No new charge is created during rubbing; only transfer occurs.
• Transferred electrons are a small fraction of the total electrons in the material.

Conductors and Insulators

• Conductors allow electricity to pass through easily; insulators do not.
• Conductors have free-moving electric charges (electrons).
• Metals, human and animal bodies, and earth are conductors.
• Non-metals like glass, porcelain, plastic, nylon, and wood are insulators.
• Semiconductors have intermediate resistance.
• Charge transferred to a conductor distributes over the entire surface.
• Charge put on an insulator stays at one place.
• Nylon or plastic combs electrify on combing dry hair, but metal spoons do not.
• Charges on metal leak through the body to the ground because both are conductors.
• A metal rod with a wooden or plastic handle shows charging signs when rubbed without touching the metal part.

Basic Properties of Electric Charge

• There are two types of charges: positive and negative, which can cancel each other's effects.
• Point charges are charged bodies with sizes much smaller than the distances between them.

Additivity of Charges

• The total charge of a system of point charges is the algebraic sum of individual charges q = q1 + q2 + q3 + \ldots + qn.
• Charge has magnitude but no direction, similar to mass.
• Mass is always positive, while charge can be positive or negative.
• Proper signs must be used when adding charges.
• Example: A system with charges +1, +2, –3, +4, and –5 has a total charge of -1.

Charge is Conserved

• When bodies are charged by rubbing, electrons transfer, but no new charges are created or destroyed.
• In an isolated system, charge redistribution may occur, but the total charge remains constant.
• Conservation of charge is experimentally established.
• A neutron can turn into a proton and an electron, creating equal and opposite charges, so the total charge remains zero.

Quantisation of Charge

• All free charges are integral multiples of the basic unit of charge e.
• Charge q on a body is q = ne, where n is an integer.
• e is the charge of an electron or proton.
• The charge on an electron is -e, and the charge on a proton is +e.
• The quantization of charge was suggested by Faraday's laws of electrolysis and demonstrated by Millikan.
• In SI units, the unit of charge is the coulomb (C).
• One coulomb is the charge flowing through a wire in 1 second if the current is 1 ampere (A).
• The value of e is 1.602192 \times 10^{-19} C.
• A charge of -1 C contains about 6 \times 10^{18} electrons.
• Smaller units like microcoulomb (\muC) and millicoulomb (mC) are used.
• If a body contains n1 electrons and n2 protons, the total charge is (n2 - n1)e.
• At the macroscopic level, charge appears continuous because the step size e is very small.
• At the microscopic level, charges appear in discrete lumps, and quantization cannot be ignored.

Example 1.1

• If 10^9 electrons move out of a body every second, the time required to get a total charge of 1 C on another body is:
  • Charge given out in one second is 1.6 \times 10^{-19} \times 10^9 C = 1.6 \times 10^{-10} C.
  • Time required = 1 \div (1.6 \times 10^{-10}) s = 6.25 \times 10^9 s = 198 years.

Example 1.2

• To calculate the positive and negative charge in a cup of water (250 g):
  • One mole (18 g) of water contains 6.02 \times 10^{23} molecules.
  • Number of molecules = (250/18) \times 6.02 \times 10^{23}.
  • Each molecule has 10 electrons and 10 protons.
  • Total charge = (250/18) \times 6.02 \times 10^{23} \times 10 \times 1.6 \times 10^{-19} C = 1.34 \times 10^7 C.

Coulomb's Law

• Coulomb's law quantifies the force between two point charges.
• The force varies inversely as the square of the distance and directly proportionally to the product of the magnitudes of the charges.
• The force acts along the line joining the two charges.
• If two point charges q1 and q2 are separated by a distance r in a vacuum, the force F is:
  • F = k \frac{q1 q2}{r^2}, where k is a constant
• Coulomb used a torsion balance to measure the force between charged spheres.
• To determine the relationship, Coulomb used identical uncharged spheres to halve the charge on a sphere by contact sharing. Repeating this process, charges q/2, q/4 etc. were achieved.
• Varying the distance for fixed charges and varying the charges at a fixed distance, Coulomb determined the relationship, Eq. (1.1).
• Coulomb's law is used to define the unit of charge. The choice of k determines the unit of charge.
• In SI units, k \approx 9 \times 10^9 Nm^2C^{-2}. Using this k, the unit of charge is the coulomb, C.
• 1 C is the charge that experiences a repulsive force of 9 \times 10^9 N when placed 1 m apart from an identical charge in a vacuum.
• One coulomb is too large a unit for electrostatics, so smaller units like mC or \mu C are used.
• The constant k is k=\frac{1}{4\pi \epsilon0}, so F = \frac{1}{4\pi \epsilon0} \frac{q1 q2}{r^2}
  • \epsilon_0 is the permittivity of free space with a value of 8.854 \times 10^{-12} C^2 N^{-1} m^{-2}.
• In vector notation:
  • \vec{r}{21} = \vec{r}2 - \vec{r}_1 (vector from 1 to 2)
  • \vec{r}{12} = \vec{r}1 - \vec{r}2 = -\vec{r}{21}
  • \hat{r}{21} = \frac{\vec{r}{21}}{r}
  • \mathbf{F}{21} = \frac{1}{4 \pi \epsilon0} \frac{q1 q2}{r{21}^2} \hat{r}{21}

Remarks on Coulomb’s Law

• Valid for any sign of q1 and q2.
• Like signs: \mathbf{F}{21} is along \hat{r}{21} (repulsion).
• Opposite signs: \mathbf{F}{21} is along -\hat{r}{21} (attraction).
• Coulomb’s law agrees with Newton’s third law: \mathbf{F}{12} = -\mathbf{F}{21}.
• The equation gives the force between two charges in a vacuum. The explanation becomes complicated when placed in matter.

Example 1.3

Compare electrostatic and gravitational forces with:

i) for an electron and a proton

Electric force:
Fe = \frac{1}{4 \pi \epsilon0} \frac{e^2}{r^2}

Gravitational force:
FG = G \frac{me m_p}{r^2}

Ratio:
\frac{Fe}{FG} = \frac{e^2}{4 \pi \epsilon0 G me m_p} \approx 2.4 \times 10^{39}

ii) for two protons

Ratio:
\frac{Fe}{FG} = \\frac{e^2}{4 \pi \epsilon0 G mp^2} \approx 1.3 \times 10^{36}\

Significance: electric forces are considerably stronger than gravitational forces.

(b) Acceleration of electron and proton

Distance between them : r = 1 \text{\AA} = 10^{-10} \, \text{m}

Force between electron and proton:
|F| = \frac{1}{4 \pi \epsilon_0} \frac{e^2}{r^2} = 8.987 \times 10^9 \frac{(1.6 \times 10^{-19})^2}{(10^{-10})^2} \approx 2.3 \times 10^{-8} \, \text{N}

Acceleration of electron
ae = \frac{F}{me} = \frac{2.3 \times 10^{-8}}{9.11 \times 10^{-31}} \approx 2.5 \times 10^{22} \, \text{m/s}^2

Acceleration of proton
ap = \frac{F}{mp} = \frac{2.3 \times 10^{-8}}{1.67 \times 10^{-27}} \approx 1.4 \times 10^{19} \, \text{m/s}^2

Explanation: The effect of gravity is negligible compared to Coulomb force.

Example 1.4

A charged sphere A suspended by a nylon thread. Another charged metallic sphere B held by an insulating handle is brought close to A at 10 cm apart. Identify repulsion of A. Spheres A and B are touched by uncharged spheres C and D respectively. Remove C and D and bring B closer to A to 5 cm and expected repulsion of A has to be identified.

Initial force
F = \frac{1}{4 \pi \epsilon_0} \frac{qq'}{r^2}

Charges after touching: q/2 and q'/2

New force: F' = \frac{1}{4 \pi \epsilon0} \frac{(q/2)(q'/2)}{(r/2)^2} F' = \frac{1}{4 \pi \epsilon0} \frac{qq'}{r^2} = F

Forces Between Multiple Charges

• Coulomb’s law applies to mutual electric forces between two charges.
• When multiple charges surround a charge, the net force is the vector sum of individual forces.
• Individual forces remain unaffected by other charges.
• This principle is known as superposition.

Force on q1 due to q2 and q_3

(a)Three charges

Force of q2 on q1:
\mathbf{F}{12} = \frac{1}{4 \pi \epsilon0} \frac{q1 q2}{r{12}^2} \hat{r}{12}

Force of q3 on q1:
\mathbf{F}{13} = \frac{1}{4 \pi \epsilon0} \frac{q1 q3}{r{13}^2} \hat{r}{13}

Resultant force on q1:
\mathbf{F}1 = \mathbf{F}{12} + \mathbf{F}_{13}

(b) Multiple charges

Total Force:
\mathbf{F}1 = \sum{i=2}^{n} \mathbf{F}_{1i}

General formula:
\mathbf{F}1 = \frac{1}{4 \pi \epsilon0} \sum{i=2}^{n} \frac{q1 qi}{r{1i}^2} \hat{r}{1i} where: \mathbf{F}{1i}​ is force on q1​ due to qi
r​1i​ is the distance between q1​ and qi
\hat{r}_{1i}
unit vector pointing from qi​ to q1

Example 1.5

Charges q1, q2, q3 each equal to q at the vertices of an equilateral triangle of side l. A Charge Q with the same sign as q is placed at the centroid O of the triangle.

• The distance from each vertex to the centroid is \frac{l}{\sqrt{3}}.
• The forces on Q due to each charge q are equal in magnitude:

F1 = F2 = F3 = \frac{1}{4 \pi \epsilon0} \frac{qQ}{(\frac{l}{\sqrt{3}})^2} = \frac{3}{4 \pi \epsilon_0} \frac{qQ}{l^2}

The forces are directed along AO, BO, and CO respectively. Due to the symmetry the forces balance out and the total force is 0.
\mathbf{F}_{\text{total}} = 0

Example 1.6

Charges q, q and -q at the vertices of an equilateral triangle with side length l. Forces on each charge has to be identified.

Force on q at A due to q at B:
F{AB} = \frac{1}{4 \pi \epsilon0} \frac{q^2}{l^2}

Force on q at A due to -q at C:
F{AC} = \frac{1}{4 \pi \epsilon0} \frac{q^2}{l^2}

Resultant on q at A:
FA = \frac{1}{4 \pi \epsilon0} \frac{q^2}{l^2}

Electric Field

Definition

• The electric field at a point is the force per unit charge that a positive test charge would experience if placed at that point.
• The field is created by a source charge.
• If a test charge q is placed at a point where the electric field is \mathbf{E}, the force on the test charge is \mathbf{F}=q\mathbf{E}.
• The SI unit of electric field is N/C (Newton per Coulomb).

Electric Field E due to a charge Q at a distance r:
\mathbf{E} = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2} \hat \mathbf{r}

Where:
\epsilon_0 = permittivity of free space
r = distance from the charge Q to the point
\hat \mathbf{r} = unit vector pointing from the charge Q to the point

If a point charge q is brought at any point around Q, Q will experience a force due to q. Making this test charge q negligibly small solves this problem.
E = \lim_{q \to 0} \frac{F}{q}

Electric Field due to a System of Charges

\mathbf{E}(\mathbf{r}) = \frac{1}{4 \pi \epsilon0} \sum{i=1}^{n} \frac{qi}{ri^2} \hat{\mathbf{r}}_i

Where:
\mathbf{E}(\mathbf{r}) is the total electric field at point r
qi​ are the charges
r​i are the distances from charges qi​ to the point
\hat{\mathbf{r}}_i unit vectors pointing from the charges qi​ to the point

Physical Significance of Electric Field

• Electric field is a characteristic of the system of charges and is independent of the test charge that you place at a point to determine the field.
• The electric field is a vector field, since force is a vector quantity.
• Accelerated motion of charge q1 produces electromagnetic waves, which then propagate with the speed c, reach q2 and cause a force on q2.
• Electric and magnetic fields have an independent dynamics of their own, i.e., they evolve according to laws of their own.
• They can also transport energy.

Example 1.7

An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude 2.0 \times 10^4 N C^{-1}.
The field is upward, so the negatively charged electron experiences a downward force of magnitude eE. The acceleration of the electron is
ae = \frac{eE}{me}
Starting from rest, the time required by the electron to fall through a distance h is given by te = \sqrt{\frac{2h}{ae}},
te = \sqrt{\frac{2hme}{eE}} = \sqrt{\frac{2 \times 1.5 \times 10^{-2} \times 9.11 \times 10^{-31}}{1.6 \times 10^{-19} \times 2.0 \times 10^4}} \approx 2.9 \times 10^{-9} \, \text{s}

Time of fall for the proton
tp = \sqrt{\frac{2hmp}{eE}} \approx 1.3 \times 10^{-7} \, \text{s}
The heavier particle (proton) takes a greater time to fall through the same distance.

Example 1.8

Two point charges q1 and q2
q1 = +10^{-8} C q2 = -10^{-8} C
are placed 0.1 \, \text{m} apart. Calculate the electric fields at points A, B and C shown in Fig. 1.11.

Electric Field Lines

Properties

• Electric field lines are a way of pictorially mapping the electric field around a configuration of charges.
• The tangent to a field line at each point is in the direction of the net field at that point.
• The magnitude of the field is represented by the density of field lines.
• Field lines start from positive charges and end at negative charges.
• In a charge-free region, electric field lines can be taken to be continuous curves without any breaks.
• Two field lines can never cross each other.
• Electrostatic field lines do not form any closed loops.

Electric Flux

\Delta \phi = \mathbf{E} \cdot \Delta \mathbf{S} = E \Delta S \cos \theta

Where:
Δϕ is the electric flux through the area element
E is the electric field at the location of the area element
ΔS is the area element
θ is the angle between the electric field and the normal to the area element

For a closed surface S:
\Phi \approx \sum \mathbf{E} \cdot \Delta \mathbf{S}

Electric Dipole

Definition

An electric dipole is a pair of equal and opposite point charges q and –q, separated by a distance 2a.

i) The field of electric dipole. For points on the axis

Electric Field E:
E = \frac{q}{4 \pi \epsilon_0} [\frac{1}{(r-a)^2} - \frac{1}{(r+a)^2}]

For r>>a E:
E=\frac{4qa}{4\pi \epsilon_0 r^3}

ii) For points on the equatorial plane

The magnitudes of the electric fields due to the two charges +q and –q:
E{+q} = \frac{q}{4 \pi \epsilon0 (r^2 + a^2)}
E{-q} = \frac{q}{4 \pi \epsilon0 (r^2 + a^2)}

Total electric field
E = \frac{-2qa}{4 \pi \epsilon_0 (r^2 + a^2)^{3/2}}

For r >> a, this reduces to
E = \frac{-qa}{4 \pi \epsilon_0 r^3}

Definition of dipole moment

\mathbf{p} = q \times 2a \hat{\mathbf{p}}

At a point on the dipole axis
E = \frac{2\mathbf{p}}{4 \pi \epsilon_0 r^3}

At a point on the equatorial plane
E = \frac{-\mathbf{p}}{4 \pi \epsilon_0 r^3}

Example 1.9

For two charges q = \pm 10^{-5} C
are Placed 5.0 \, \text{mm} apart.
Point P is at on the axis is 15 \, \text{cm} away from its centre O on the side of the positive charge. Point Q, is 15 \, \text{cm} away from O on a line passing through O and normal to the axis of the dipole.

Dipole in a Uniform External Field

• If a dipole is placed in a uniform external electric field, it experiences a torque.
• \tau = pE \sin \theta
• \vec \tau = \vec p \times \vec E
• This torque will tend to align the dipole with the field \mathbf{E}.
• When \mathbf{p} is aligned with \mathbf{E}, the torque is zero.

Continuous Charge Distribution

• Continuous charge distributions are described in terms of charge density.
  Linear charge density:
  \lambda = \frac{\Delta Q}{\Delta l}
  Surface charge density:
  \sigma = \frac{\Delta Q}{\Delta S}
  Volume charge density:
  \rho = \frac{\Delta Q}{\Delta V}

Electric field
\mathbf{E}(\mathbf{r}) = \frac{1}{4 \pi \epsilon_0} \int \frac{\rho \, dV}{r^2} \hat{\mathbf{r}}

Gauss’s Law

\Phi = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q{enc}}{\epsilon0}

Where:
Φ is the electric flux
E is the electric field
dA is an infinitesimal area vector of the surface, pointing outwards
Qenc​ is the total charge enclosed by the surface
ϵ0​ is the permittivity of free space

Applications of Gauss’s Law

Field due to an infinitely long straight uniformly charged wire

• Using a cylindrical Gaussian surface:
  E = \frac{\lambda}{2 \pi \epsilon_0 r}

Field due to a uniformly charged infinite plane sheet

E = \frac{\sigma}{2 \epsilon_0}

Field due to a uniformly charged thin spherical shell

Outside the shell (r > R):
E = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}

Inside the shell (r < R):
E = 0

Example 1.12
Charge density: \displaystyle \rho = -\frac{3Ze}{4 \pi R^3}

(i)\mathbf{r} < R gives, \mathbf{E} = \frac{Ze}{4 \pi \epsilon0} \frac{1}{r^2} - \frac{Ze}{4 \pi \epsilon0} \frac{r}{R^3} (ii) \mathbf{r} > R:
\E=0$$