Electric Charges and Fields
Electricity begins with a simple observation: certain materials attract each other when rubbed.
Start with the simplest version: this lesson is about Electric Charges and Fields. If you can explain the core idea to a friend using everyday language, examples, and one clear reason why it matters, you have moved from memorising to understanding.
Electricity begins with a simple observation: certain materials attract each other when rubbed. This ancient discovery laid the foundation for understanding how everything in the universe communicates through invisible force fields. Electric charges are fundamental properties of matter that either repel or attract each other, and the invisible territory between them is called the electric field—imagine it as an invisible ocean of force that surrounds every charged particle, ready to influence anything that ventures into it.
Building from Class 11 Knowledge
In Class 11, you learned about forces and fields in the context of gravity. Gravity is always attractive, but electricity is more interesting: it can push or pull. While a mass always attracts other masses, a charged object can either attract or repel another charged object. The electric field electrostatic-potential-and-capacitance works similarly to a gravitational field, but with this crucial difference—it has both positive and negative aspects.
The Concept of Electric Charge
Electric charge is what makes electricity happen. There are two types: positive charge (like that on a glass rod rubbed with silk) and negative charge (like that on a plastic rod rubbed with fur). The fundamental rule is simple: like charges repel, opposite charges attract. This isn't mysterious—it's just how nature works at the smallest scales.
The smallest possible charge is carried by an electron (negative) or a proton (positive): e = 1.6 × 10⁻¹⁹ coulombs. All charges in nature are integer multiples of this elementary charge. This is called charge quantization.
Electric Force and Coulomb's Law
Two charges interact through a force described by Coulomb's Law:
F = k(q₁ × q₂)/r²
Where:
- q₁ and q₂ are the charges
- r is the distance between them
- k = 9 × 10⁹ N·m²/C² (Coulomb's constant)
Think of it this way: the force gets weaker as charges move apart, following an inverse square law—just like how light from a candle dims as you walk away from it.
Understanding Electric Field
An electric field is the influence that a charge creates in the space around it. When you place a test charge in this field, it experiences a force. The electric field strength (E) tells us the force per unit charge:
E = F/q = k(Q)/r²
Imagine you're standing at various distances from a loudspeaker. The sound (field) gets weaker farther away. Similarly, electric fields get weaker with distance, but they extend infinitely—they just become very small far away.
Superposition Principle
When multiple charges create fields in the same region, their effects add up. This is the principle of superposition. If you have three charges, each creates its own field at a point, and the total field is simply the vector sum of all three individual fields. This is powerful because it lets us solve complex problems by breaking them into simpler pieces.
Key Concepts Summary
- Charges are quantized (come in multiples of electron charge)
- Like charges repel, opposite charges attract
- Electric force follows Coulomb's law (inverse square)
- Electric field is a property of space around a charge
- Fields from multiple charges superpose
Related Topics
electrostatic-potential-and-capacitance | current-electricity | moving-charges-and-magnetism
Socratic Questions
- If electric charge were continuous (not quantized) rather than coming in discrete packets like electrons, how would atoms and molecules be fundamentally different?
- Why is it impossible to have a magnetic monopole (an isolated north or south pole), yet we can isolate positive and negative electric charges separately?
- If the inverse-square law of Coulomb's law were slightly different—say, inverse-cube—how would atomic structure and chemical bonding change?
- When you comb your hair and it stands on end, the forces involved are electric. Why don't gravitational forces (which exist between all masses) achieve the same effect?
- Can you imagine a scenario where the principle of superposition would fail? What would the universe need to be like for fields not to add together simply?
NCERT Textbook Content
1.1 Introduction
All of us have the experience of seeing a spark or hearing a crackle when we take off our synthetic clothes or sweater, particularly in dry weather. Have you ever tried to find any explanation for this phenomenon? Another common example of electric discharge is the lightning that we see in the sky during thunderstorms. We also experience a sensation of an electric shock either while opening the door of a car or holding the iron bar of a bus after sliding from our seat. The reason for these experiences is discharge of electric charges through our body, which were accumulated due to rubbing of insulating surfaces.
> [!info] Definition > Electrostatics deals with the study of forces, fields and potentials arising from static charges.
1.2 Electric Charge
Historically the credit of discovery of the fact that amber rubbed with wool or silk cloth attracts light objects goes to Thales of Miletus, Greece, around 600 BC. The name electricity is coined from the Greek word elektron meaning amber.
Observations from Experiments
- Two glass rods rubbed with wool or silk cloth repel each other
- Two plastic rods rubbed with cat's fur repel each other
- A plastic rod attracts a glass rod
- The glass rod and wool attract each other
These observations led to the conclusion that there are only two kinds of electric charge:
> [!important] Fundamental Rule > - Like charges repel each other > - Unlike charges attract each other
The property which differentiates the two kinds of charges is called the polarity of charge.
By convention (established by Benjamin Franklin):
- Charge on glass rod or cat's fur = Positive (+)
- Charge on plastic rod or silk = Negative (−)
1.3 Conductors and Insulators
| Type | Definition | Examples |
|---|---|---|
| Conductors | Substances that readily allow passage of electricity | Metals, human body, earth |
| Insulators | Substances that offer high resistance to electricity | Glass, porcelain, plastic, nylon, wood |
| Semiconductors | Intermediate resistance between conductors and insulators | Silicon, Germanium |
> [!note] Key Difference > - In conductors: charge distributes over the entire surface > - In insulators: charge stays at the same place where it was put
1.4 Basic Properties of Electric Charge
1.4.1 Additivity of Charges
Charges add up like real numbers (scalars). If a system contains n charges $q_1, q_2, q_3, \ldots, q_n$, then the total charge is:
$$Q_{total} = q_1 + q_2 + q_3 + \ldots + q_n$$
> [!example] Example > Total charge of a system containing charges +1, +2, −3, +4 and −5 (in arbitrary units): > $(+1) + (+2) + (-3) + (+4) + (-5) = -1$
1.4.2 Conservation of Charge
> [!important] Law of Conservation of Charge > The total charge of an isolated system is always conserved. Charge can neither be created nor destroyed, only transferred from one body to another.
When we rub two bodies, what one body gains in charge, the other body loses.
Example: When a neutron turns into a proton and an electron:
- Before: Total charge = 0 (neutron is neutral)
- After: Total charge = (+e) + (−e) = 0
1.4.3 Quantisation of Charge
All free charges are integral multiples of a basic unit of charge denoted by e:
$$q = ne$$
where $n$ is any integer (positive or negative).
Value of elementary charge: $$e = 1.602192 \times 10^{-19} \text{ C}$$
> [!info] Practical Units > - 1 μC (micro coulomb) = $10^{-6}$ C > - 1 mC (milli coulomb) = $10^{-3}$ C > - 1 C contains about $6 \times 10^{18}$ electrons
Example 1.1
Problem: If $10^9$ electrons move out of a body every second, how much time is required to get a total charge of 1 C on the other body?
Solution:
- Charge given out per second = $1.6 \times 10^{-19} \times 10^9$ C = $1.6 \times 10^{-10}$ C
- Time required = $\frac{1 \text{ C}}{1.6 \times 10^{-10} \text{ C/s}} = 6.25 \times 10^9$ s ≈ 198 years
Example 1.2
Problem: How much positive and negative charge is there in a cup of water?
Solution:
- Mass of water = 250 g
- Number of molecules = $\frac{250}{18} \times 6.02 \times 10^{23}$
- Each H₂O molecule has 10 electrons and 10 protons
- Total charge = $\frac{250}{18} \times 6.02 \times 10^{23} \times 10 \times 1.6 \times 10^{-19}$ C = 1.34 × 10⁷ C
1.5 Coulomb's Law
Coulomb's law is a quantitative statement about the force between two point charges.
> [!important] Coulomb's Law > The magnitude of the electrostatic force between two point charges $q_1$ and $q_2$ separated by distance $r$ in vacuum is: > $$F = k\frac{|q_1 q_2|}{r^2}$$
where $k = \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9$ N·m²/C²
Permittivity of free space: $$\varepsilon_0 = 8.854 \times 10^{-12} \text{ C}^2 \text{ N}^{-1}\text{m}^{-2}$$
Vector Form of Coulomb's Law
$$\vec{F}_{21} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r_{21}^2} \hat{r}_{21}$$
where $\hat{r}_{21}$ is the unit vector from $q_1$ to $q_2$.
> [!note] Important Points > - The equation handles both like and unlike charges automatically > - Coulomb's law agrees with Newton's third law: $\vec{F}_{12} = -\vec{F}_{21}$
Example 1.3
Comparing Electric and Gravitational Forces
For an electron and proton: $$\frac{F_e}{F_G} = \frac{e^2}{4\pi\varepsilon_0 G m_p m_e} = 2.4 \times 10^{39}$$
> [!important] Key Insight > Electric forces are enormously stronger than gravitational forces (by a factor of ~10³⁹)!
1.6 Forces Between Multiple Charges
Principle of Superposition
> [!important] Superposition Principle > The force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time.
For a system of charges $q_1, q_2, \ldots, q_n$, the total force on $q_1$ is:
$$\vec{F}_1 = \vec{F}_{12} + \vec{F}_{13} + \ldots + \vec{F}_{1n} = \frac{q_1}{4\pi\varepsilon_0} \sum_{i=2}^{n} \frac{q_i}{r_{1i}^2} \hat{r}_{1i}$$
Example 1.5
Three Equal Charges at Vertices of Equilateral Triangle
For three charges $q$ at vertices of an equilateral triangle with side $l$, the force on charge $Q$ at the centroid:
$$\vec{F} = 0$$
By symmetry, the three forces sum to zero.
Example 1.6
Charges q, q, and −q at Triangle Vertices
The forces on each charge have magnitude: $$F = \frac{q^2}{4\pi\varepsilon_0 l^2}$$
The resultant forces are:
- $|\vec{F}_1| = |\vec{F}_2| = F$
- $|\vec{F}_3| = \sqrt{3}F$
1.7 Electric Field
> [!definition] Electric Field > The electric field $\vec{E}$ at a point is defined as the force experienced by a unit positive test charge placed at that point: > $$\vec{E} = \lim_{q \to 0} \frac{\vec{F}}{q}$$
For a point charge $Q$: $$\vec{E}(\vec{r}) = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2} \hat{r}$$
SI Unit: N/C or V/m
Electric Field Due to a System of Charges
$$\vec{E}(\vec{r}) = \frac{1}{4\pi\varepsilon_0} \sum_{i=1}^{n} \frac{q_i}{r_{iP}^2} \hat{r}_{iP}$$
Physical Significance of Electric Field
The concept of electric field is crucial for understanding:
- Time-dependent electromagnetic phenomena
- How effects propagate at the speed of light
- The independent dynamics of electromagnetic fields
- Energy transport by electromagnetic waves
Example 1.7
Electron vs Proton Falling in Electric Field
For $E = 2.0 \times 10^4$ N/C and $h = 1.5$ cm:
- Electron: $t_e = \sqrt{\frac{2hm_e}{eE}} = 2.9 \times 10^{-9}$ s
- Proton: $t_p = \sqrt{\frac{2hm_p}{eE}} = 1.3 \times 10^{-7}$ s
The proton takes longer because it's heavier (unlike free fall under gravity where time is independent of mass).
Example 1.8
Electric Field of a Dipole
For charges ±10 μC placed 5.0 mm apart:
(a) On the axis at 15 cm: $E_A = 7.2 \times 10^4$ N/C (toward the right)
(b) On equatorial plane at 15 cm: $E_C = 9 \times 10^3$ N/C (toward the right)
1.8 Electric Field Lines
Electric field lines are curves drawn such that the tangent at each point gives the direction of the electric field.
> [!important] Properties of Field Lines > 1. Field lines start from positive charges and end at negative charges > 2. Field lines are continuous curves without breaks > 3. Two field lines never cross each other > 4. Field lines do not form closed loops (electrostatic fields are conservative) > 5. Density of field lines indicates field strength
1.9 Electric Flux
> [!definition] Electric Flux > The electric flux through an area element $\Delta\vec{S}$ is: > $$\Delta\phi = \vec{E} \cdot \Delta\vec{S} = E \Delta S \cos\theta$$
where $\theta$ is the angle between $\vec{E}$ and the normal to the surface.
SI Unit: N·m²/C or V·m
1.10 Electric Dipole
An electric dipole is a pair of equal and opposite point charges $q$ and $-q$ separated by distance $2a$.
Dipole Moment
$$\vec{p} = q \times 2a \, \hat{p}$$
Direction: from $-q$ to $+q$
SI Unit: C·m
Electric Field of a Dipole
On the axis (at distance $r$ from center, $r >> a$): $$\vec{E} = \frac{2\vec{p}}{4\pi\varepsilon_0 r^3}$$
On the equatorial plane (at distance $r$ from center, $r >> a$): $$\vec{E} = -\frac{\vec{p}}{4\pi\varepsilon_0 r^3}$$
> [!note] Key Point > Dipole field falls off as $1/r^3$ (faster than the $1/r^2$ dependence for a point charge)
Example 1.9
Dipole Field Calculations
For dipole moment $p = 5 \times 10^{-8}$ C·m:
(a) On axis at 15 cm: $E = 2.6 \times 10^5$ N/C
(b) On equatorial plane at 15 cm: $E = 1.33 \times 10^5$ N/C
1.11 Dipole in a Uniform External Field
For a dipole with moment $\vec{p}$ in uniform field $\vec{E}$:
Net Force: $\vec{F} = 0$ (forces on $+q$ and $-q$ cancel)
Torque: $$\vec{\tau} = \vec{p} \times \vec{E}$$ $$|\tau| = pE\sin\theta$$
The torque tends to align the dipole with the field.
> [!note] Non-uniform Field > In a non-uniform field, the dipole experiences both torque AND net force in the direction of increasing field (when $\vec{p}$ parallel to $\vec{E}$).
1.12 Continuous Charge Distribution
For continuous charge distributions, we define:
| Type | Symbol | Definition | Unit |
|---|---|---|---|
| Linear charge density | $\lambda$ | $\frac{\Delta Q}{\Delta l}$ | C/m |
| Surface charge density | $\sigma$ | $\frac{\Delta Q}{\Delta S}$ | C/m² |
| Volume charge density | $\rho$ | $\frac{\Delta Q}{\Delta V}$ | C/m³ |
The electric field due to continuous distribution: $$\vec{E} \approx \frac{1}{4\pi\varepsilon_0} \sum_{all \, \Delta V} \frac{\rho \Delta V}{r'^2} \hat{r}'$$
1.13 Gauss's Law
> [!important] Gauss's Law > The total electric flux through any closed surface $S$ equals $1/\varepsilon_0$ times the total charge enclosed: > $$\phi = \oint \vec{E} \cdot d\vec{S} = \frac{q_{enclosed}}{\varepsilon_0}$$
Important Points about Gauss's Law
- Valid for any closed surface (any shape or size)
- $q$ includes sum of all charges inside the surface
- $\vec{E}$ is due to all charges (inside and outside)
- The Gaussian surface should not pass through discrete charges
- Most useful when the system has symmetry
- Based on inverse square law of Coulomb's law
Example 1.10
Flux Through a Cube
For $E_x = \alpha x^{1/2}$ with $\alpha = 800$ N/C·m^(1/2) and cube side $a = 0.1$ m:
(a) Flux: $\phi = \alpha a^{5/2}(\sqrt{2}-1) = 1.05$ N·m²/C
(b) Charge inside: $q = \varepsilon_0 \phi = 9.27 \times 10^{-12}$ C
Example 1.11
Flux Through a Cylinder
For uniform field $E = 200$ N/C, cylinder length 20 cm, radius 5 cm:
- Through each flat face: $\phi = 1.57$ N·m²/C
- Through curved surface: $\phi = 0$
- Net flux: $\phi = 3.14$ N·m²/C
- Net charge: $q = 2.78 \times 10^{-11}$ C
1.14 Applications of Gauss's Law
1.14.1 Infinitely Long Straight Uniformly Charged Wire
$$\vec{E} = \frac{\lambda}{2\pi\varepsilon_0 r} \hat{n}$$
where $\lambda$ is linear charge density and $\hat{n}$ is radial unit vector.
1.14.2 Uniformly Charged Infinite Plane Sheet
$$\vec{E} = \frac{\sigma}{2\varepsilon_0} \hat{n}$$
where $\sigma$ is surface charge density.
> [!note] Field is Independent of Distance > The field is constant and does not depend on distance from the sheet!
1.14.3 Uniformly Charged Thin Spherical Shell
Outside the shell ($r \geq R$): $$\vec{E} = \frac{q}{4\pi\varepsilon_0 r^2} \hat{r}$$
Inside the shell ($r < R$): $$\vec{E} = 0$$
> [!important] Key Result > For points outside, the shell behaves as if all charge is concentrated at the center. Inside the shell, the field is zero.
Example 1.12
Early Atomic Model (Thomson Model)
For an atom with positive point charge $Ze$ at nucleus surrounded by uniform negative charge density up to radius $R$:
For $r < R$: $$E(r) = \frac{Ze}{4\pi\varepsilon_0}\left(\frac{1}{r^2} - \frac{r}{R^3}\right)$$
For $r > R$: $$E(r) = 0$$ (atom is neutral)
Summary
- Electric charge has two types (positive and negative); like charges repel, unlike attract.
- Conductors allow free movement of charges; insulators do not.
- Additivity: charges add algebraically - Conservation: total charge is constant in isolated system - Quantisation: $q = ne$ where $e = 1.6 \times 10^{-19}$ C
- Properties of charge:
- Coulomb's Law: $F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}$
- Superposition Principle: Forces add vectorially
- Electric Field: $\vec{E} = \vec{F}/q$ (force per unit charge)
- Electric Field Lines: Start from positive, end at negative charges; density indicates field strength
- Electric Dipole: $\vec{p} = q \times 2a$; field falls as $1/r^3$
- Dipole in uniform field: experiences torque $\vec{\tau} = \vec{p} \times \vec{E}$
- Electric Flux: $\phi = \vec{E} \cdot \vec{A} = EA\cos\theta$
- Gauss's Law: $\oint \vec{E} \cdot d\vec{S} = q_{enclosed}/\varepsilon_0$
- Infinite wire: $E = \lambda/(2\pi\varepsilon_0 r)$ - Infinite sheet: $E = \sigma/(2\varepsilon_0)$ - Spherical shell: $E = q/(4\pi\varepsilon_0 r^2)$ outside, $E = 0$ inside
- Applications of Gauss's Law:
Exercises
1.1 What is the force between two small charged spheres having charges of $2 \times 10^{-7}$ C and $3 \times 10^{-7}$ C placed 30 cm apart in air?
1.2 The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge −0.8 μC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?
1.3 Check that the ratio $ke^2/Gm_em_p$ is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?
1.4 (a) Explain the meaning of the statement 'electric charge of a body is quantised'. (b) Why can one ignore quantisation of electric charge when dealing with macroscopic charges?
1.5 When a glass rod is rubbed with a silk cloth, charges appear on both. Explain how this observation is consistent with the law of conservation of charge.
1.6 Four point charges $q_A = 2$ μC, $q_B = -5$ μC, $q_C = 2$ μC, and $q_D = -5$ μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?
1.7 (a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not? (b) Explain why two field lines never cross each other at any point?
1.8 Two point charges $q_A = 3$ μC and $q_B = -3$ μC are located 20 cm apart in vacuum. (a) What is the electric field at the midpoint O of the line AB? (b) If a negative test charge of magnitude $1.5 \times 10^{-9}$ C is placed at this point, what is the force experienced?
1.9 A system has two charges $q_A = 2.5 \times 10^{-7}$ C and $q_B = -2.5 \times 10^{-7}$ C located at A(0, 0, −15 cm) and B(0, 0, +15 cm). What are the total charge and electric dipole moment of the system?
1.10 An electric dipole with dipole moment $4 \times 10^{-9}$ C·m is aligned at 30° with a uniform electric field of magnitude $5 \times 10^4$ N/C. Calculate the magnitude of the torque acting on the dipole.
1.11 A polythene piece rubbed with wool is found to have a negative charge of $3 \times 10^{-7}$ C. (a) Estimate the number of electrons transferred. (b) Is there a transfer of mass from wool to polythene?
1.12 (a) Two insulated charged copper spheres A and B have their centres separated by 50 cm. What is the mutual force of repulsion if the charge on each is $6.5 \times 10^{-7}$ C? (b) What is the force if each sphere is charged double the above amount, and the distance between them is halved?
1.13 Consider tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?
1.14 Consider a uniform electric field $\vec{E} = 3 \times 10^3 \hat{i}$ N/C. (a) What is the flux through a square of 10 cm side parallel to the yz plane? (b) What is the flux if the normal makes 60° with the x-axis?
1.15 What is the net flux of the uniform electric field of Exercise 1.14 through a cube of side 20 cm oriented with faces parallel to the coordinate planes?
1.16 The net outward flux through the surface of a black box is $8.0 \times 10^3$ N·m²/C. (a) What is the net charge inside? (b) If the flux were zero, could you conclude there were no charges inside? Why?
1.17 A point charge +10 μC is 5 cm directly above the centre of a square of side 10 cm. What is the magnitude of the electric flux through the square?
1.18 A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
1.19 A point charge causes an electric flux of $-1.0 \times 10^3$ N·m²/C through a spherical Gaussian surface of 10.0 cm radius. (a) If the radius were doubled, how much flux would pass? (b) What is the value of the point charge?
1.20 A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre is $1.5 \times 10^3$ N/C pointing radially inward, what is the net charge on the sphere?
1.21 A uniformly charged conducting sphere of 2.4 m diameter has surface charge density 80.0 μC/m². (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface?
1.22 An infinite line charge produces a field of $9 \times 10^4$ N/C at a distance of 2 cm. Calculate the linear charge density.
1.23 Two large, thin metal plates are parallel with surface charge densities of opposite signs and magnitude $17.0 \times 10^{-22}$ C/m² on their inner faces. What is E: (a) in the outer region of the first plate, (b) in the outer region of the second plate, (c) between the plates?
