Prerequisites
Before studying this topic, make sure you understand:
- Electric Charges and Coulomb’s Law — Force between charges
- Vector addition and resolution
The Hook: Invisible Force Fields
Remember in Iron Man when Tony Stark talks about “generating a local electromagnetic field”? Or in Stranger Things when Eleven creates force fields around objects? Or the lightning scene in Thor where electric energy radiates outward?
These fictional force fields are inspired by a very real physics concept: the electric field — an invisible region of influence around every charged particle. You can’t see it, but it’s there, ready to push or pull any charge that enters it!
What is an Electric Field?
An electric field is the region around a charged object where another charged object experiences a force.
In simple terms: Every charge creates an invisible “influence zone” around itself. Any other charge entering this zone will feel a push or pull.
Why Do We Need the Concept?
Without field concept: “Charge A pushes charge B” — action at a distance (spooky!)
With field concept:
- Charge A creates a field around itself
- Charge B interacts with this field at its location
- The field exerts force on B
This makes physics more intuitive — charges don’t “reach out” to each other; they interact through fields!
Electric Field Intensity
where:
- $\vec{E}$ = electric field (N/C or V/m)
- $\vec{F}$ = force on test charge
- $q_0$ = small positive test charge
Key Point: $q_0$ must be small enough that it doesn’t disturb the source charge distribution.
Electric Field Due to a Point Charge
For a point charge $Q$ at distance $r$:
$$\boxed{\vec{E} = k\frac{Q}{r^2}\hat{r}}$$Magnitude:
$$\boxed{E = k\frac{|Q|}{r^2} = \frac{1}{4\pi\varepsilon_0}\frac{|Q|}{r^2}}$$where $k = 9 \times 10^9$ N·m²/C²
Interactive Demo: Visualize Electric Field Lines
See how electric field lines emanate from positive charges and terminate on negative charges:
Direction:
- Positive charge: Field points radially outward (away from charge)
- Negative charge: Field points radially inward (toward charge)
“Positive Pushes Out, Negative Pulls In”
Think of positive charge as a repeller — its field arrows point outward, ready to push other positive charges away.
Negative charge is an attractor — its field arrows point inward, ready to pull positive charges in.
Interactive Demo: Electric Field Visualization
Explore the interactive electric field simulator below. You can add and remove charges, drag them around, and observe how the field lines and force on a test charge change in real-time:
- Add charges: Click “+q” or “-q” buttons, then click on canvas to place
- Move charges: Drag any charge to reposition it
- Test charge: The green charge shows the force direction at that point (drag it around!)
- Toggle views: Switch between field lines, vectors, and equipotential lines
- Presets: Try the preset configurations to see common field patterns
- Remove charge: Double-click on a charge to delete it
Try the different presets and observe:
- Single +q / -q: Radial field pattern - field strength decreases with distance (inverse square law)
- Dipole: Field lines start from + and end on - (curved paths, denser near charges)
- Like Charges: Repulsion pattern - field lines never cross, symmetric about midpoint
- Parallel Plates: Uniform field between plates - field is constant everywhere!
Properties of Electric Field
1. Vector Quantity
Electric field has both magnitude and direction.
For force on charge $q$ in field $\vec{E}$:
$$\boxed{\vec{F} = q\vec{E}}$$2. Depends on Source, Not on Test Charge
Field $\vec{E}$ is created by the source charge. It exists whether or not you place a test charge there!
3. Inverse Square Law
Field strength decreases with square of distance:
- At distance $r$: $E$
- At distance $2r$: $E/4$
- At distance $3r$: $E/9$
Principle of Superposition for Electric Fields
Strategy for Multi-Charge Problems:
- Find field due to each charge separately (as if others don’t exist)
- Resolve each field into x and y components
- Add components: $E_x = \sum E_{ix}$, $E_y = \sum E_{iy}$
- Find magnitude: $E = \sqrt{E_x^2 + E_y^2}$
- Find direction: $\theta = \tan^{-1}\left(\frac{E_y}{E_x}\right)$
Electric Field Lines (Lines of Force)
Electric field lines are imaginary lines that represent the electric field visually.
Properties of Field Lines
- Start from positive charges (or infinity)
- End at negative charges (or infinity)
- Never cross each other
- Tangent at any point gives field direction at that point
- Density (closeness) represents field strength
- Perpendicular to the surface of a conductor
- Perpendicular to equipotential surfaces
Why Can’t Field Lines Cross?
If two lines crossed, there would be two different field directions at that point. But field at a point is unique — a test charge can only experience force in ONE direction!
Number of Field Lines
For a charge $Q$, the number of field lines drawn is proportional to $|Q|$.
Example:
- Charge $+2q$: Draw 8 lines outward
- Charge $+q$: Draw 4 lines outward
- Charge $-q$: Draw 4 lines inward
Electric Field Patterns for Common Configurations
1. Single Positive Charge
- Lines radiate outward in all directions
- Uniformly distributed in 3D (sphere)
- Density decreases with distance
2. Single Negative Charge
- Lines converge inward from all directions
- Uniformly distributed in 3D
3. Two Equal Positive Charges
- Lines repel each other
- Symmetric pattern
- Zero field at midpoint (by symmetry)
4. Positive and Negative Charges (Dipole)
- Lines start from positive, end at negative
- Curved paths
- Denser lines near charges
5. Parallel Plates
- Uniform field between plates (straight, parallel lines)
- Field outside ≈ 0
- $E = \frac{\sigma}{\varepsilon_0}$ (inside)
Electric Field Due to Continuous Charge Distributions
For continuous distributions, sum becomes integral:
$$\vec{E} = \int d\vec{E} = \int k\frac{dq}{r^2}\hat{r}$$Linear Charge Distribution (Line of Charge)
Charge per unit length: $\lambda = \frac{dq}{dl}$
Surface Charge Distribution (Sheet of Charge)
Charge per unit area: $\sigma = \frac{dq}{dA}$
Volume Charge Distribution (Charged 3D Object)
Charge per unit volume: $\rho = \frac{dq}{dV}$
Important Results for Standard Distributions
1. Infinite Line Charge
For an infinitely long line with linear charge density $\lambda$:
$$\boxed{E = \frac{\lambda}{2\pi\varepsilon_0 r} = \frac{2k\lambda}{r}}$$Direction: Radially outward (perpendicular to line)
2. Infinite Plane Sheet
For an infinite plane with surface charge density $\sigma$:
$$\boxed{E = \frac{\sigma}{2\varepsilon_0}}$$Amazing property: Field is uniform (doesn’t depend on distance from sheet!)
3. Two Parallel Infinite Sheets (Opposite Charges)
Between plates: $E = \frac{\sigma}{\varepsilon_0}$ (fields add)
Outside plates: $E = 0$ (fields cancel)
This is the basis for parallel plate capacitors!
4. Uniformly Charged Spherical Shell (radius R, total charge Q)
Outside ($r > R$):
$$E = k\frac{Q}{r^2}$$(behaves like point charge at center)
On surface ($r = R$):
$$E = k\frac{Q}{R^2}$$Inside ($r < R$):
$$E = 0$$(zero field inside a conductor!)
5. Uniformly Charged Solid Sphere (radius R, total charge Q)
Outside ($r > R$):
$$E = k\frac{Q}{r^2}$$Inside ($r < R$):
$$\boxed{E = k\frac{Qr}{R^3} = \frac{Q}{4\pi\varepsilon_0 R^3} \cdot r}$$Field increases linearly from center!
Memory Tricks & Patterns
Mnemonic for Field Formulas
“Every Point Keeps Quitting at Radius-Squared”
$E = k \frac{Q}{r^2}$ (for point charge)
Pattern Recognition: Field Behavior
| Configuration | Field Dependence | Special Feature |
|---|---|---|
| Point charge | $E \propto \frac{1}{r^2}$ | Inverse square |
| Infinite line | $E \propto \frac{1}{r}$ | Inverse (not square!) |
| Infinite plane | $E =$ constant | Independent of distance |
| Spherical shell (inside) | $E = 0$ | Zero field |
| Solid sphere (inside) | $E \propto r$ | Linear increase |
JEE High-Yield Pattern:
- 0D (point): $E \propto \frac{1}{r^2}$
- 1D (line): $E \propto \frac{1}{r^1}$
- 2D (plane): $E \propto \frac{1}{r^0} = \text{constant}$
Dimension of source affects how field decreases!
When to Use Electric Field vs. Coulomb’s Law
Use Electric Field ($\vec{E}$) when:
- One charge is much smaller (test charge)
- Asked about field at a point (not force)
- Multiple test charges in the same field
- Dealing with field lines or field visualization
Use Coulomb’s Law directly when:
- Both charges are comparable
- Asked specifically about force between two charges
- Simple two-charge system
Common Mistakes to Avoid
Wrong: “Negative charge, so field is negative”
Correct:
- Field magnitude is always positive: $E = k\frac{|Q|}{r^2}$
- Field direction points toward negative charge
Field is a vector — magnitude and direction are separate!
Wrong: Two fields of 3 N/C and 4 N/C → total = 7 N/C
Correct: Check angle!
- If same direction: $E = 3 + 4 = 7$ N/C
- If opposite: $E = |4 - 3| = 1$ N/C
- If perpendicular: $E = \sqrt{3^2 + 4^2} = 5$ N/C
Wrong: “Large test charge, so field is large”
Correct: Field is created by source charge, independent of test charge!
$\vec{E} = \frac{\vec{F}}{q_0}$ — even if you increase $q_0$, both $\vec{F}$ and $q_0$ increase proportionally, so $\vec{E}$ stays same.
Wrong: “Conductor has charges, so field inside is large”
Correct: In electrostatic equilibrium, field inside a conductor is always zero.
Charges redistribute on the surface to make internal field zero!
Practice Problems
Level 1: Foundation (NCERT)
A point charge of $+5$ μC is placed at the origin. Find the electric field at point $(3, 4)$ m.
Solution:
Distance: $r = \sqrt{3^2 + 4^2} = 5$ m
Field magnitude:
$$E = k\frac{Q}{r^2} = 9 \times 10^9 \times \frac{5 \times 10^{-6}}{25} = 1800 \text{ N/C}$$Direction: Along line from origin to $(3, 4)$
Unit vector: $\hat{r} = \frac{3\hat{i} + 4\hat{j}}{5}$
$$\boxed{\vec{E} = 1800 \times \frac{3\hat{i} + 4\hat{j}}{5} = 1080\hat{i} + 1440\hat{j} \text{ N/C}}$$An electron is placed in a uniform electric field of $500$ N/C. Find the force on it.
Solution:
$$F = qE = (1.6 \times 10^{-19}) \times 500$$ $$\boxed{F = 8 \times 10^{-17} \text{ N}}$$Direction: Opposite to field (electron is negative)
At what distance from a $10$ μC charge is the electric field $9000$ N/C?
Solution:
$$E = k\frac{Q}{r^2}$$ $$r^2 = k\frac{Q}{E} = 9 \times 10^9 \times \frac{10 \times 10^{-6}}{9000} = 10$$ $$\boxed{r = \sqrt{10} \approx 3.16 \text{ m}}$$Level 2: JEE Main
Two charges $+4$ μC and $-4$ μC are placed 20 cm apart. Find the electric field at the midpoint.
Solution:
Midpoint is 10 cm = 0.1 m from each charge.
Field due to $+4$ μC (pointing away from it, i.e., to the right):
$$E_1 = k\frac{4 \times 10^{-6}}{(0.1)^2} = 9 \times 10^9 \times \frac{4 \times 10^{-6}}{0.01} = 3.6 \times 10^6 \text{ N/C}$$Field due to $-4$ μC (pointing toward it, i.e., also to the right):
$$E_2 = k\frac{4 \times 10^{-6}}{(0.1)^2} = 3.6 \times 10^6 \text{ N/C}$$Both fields point in the same direction (from + to -)!
$$\boxed{E_{net} = E_1 + E_2 = 7.2 \times 10^6 \text{ N/C (from + toward -)}}$$Four charges $+q$ each are placed at the corners of a square of side $a$. Find the field at the center.
Solution:
By symmetry, fields from all four charges cancel out!
Each charge creates field toward center with magnitude $E = k\frac{q}{r^2}$ where $r = \frac{a}{\sqrt{2}}$.
But opposite charges create equal and opposite fields.
$$\boxed{E_{net} = 0}$$Key Insight: Always check for symmetry first!
An electric field $\vec{E} = 200\hat{i} + 300\hat{j}$ N/C exists in a region. Find the force on a $-2$ μC charge.
Solution:
$$\vec{F} = q\vec{E} = (-2 \times 10^{-6})(200\hat{i} + 300\hat{j})$$ $$\boxed{\vec{F} = -400\hat{i} - 600\hat{j} \text{ μN} = -0.4\hat{i} - 0.6\hat{j} \text{ mN}}$$Magnitude: $F = \sqrt{0.4^2 + 0.6^2} = 0.72$ mN
Level 3: JEE Advanced
A charge $+Q$ is uniformly distributed on a thin ring of radius $R$. Find the electric field at a point on the axis at distance $x$ from the center.
Solution:
Consider a small element $dq$ on the ring. Its distance from point P:
$$r = \sqrt{R^2 + x^2}$$Field magnitude: $dE = k\frac{dq}{R^2 + x^2}$
By symmetry: Perpendicular components cancel, only axial component survives.
Axial component: $dE_x = dE \cos\theta = dE \frac{x}{\sqrt{R^2 + x^2}}$
Total field:
$$E = \int dE_x = \int k\frac{dq}{R^2 + x^2} \cdot \frac{x}{\sqrt{R^2 + x^2}}$$ $$E = \frac{kx}{(R^2 + x^2)^{3/2}} \int dq = \frac{kx}{(R^2 + x^2)^{3/2}} \cdot Q$$ $$\boxed{E = \frac{kQx}{(R^2 + x^2)^{3/2}}}$$Special cases:
- At center ($x = 0$): $E = 0$ (by symmetry)
- Far away ($x >> R$): $E \approx \frac{kQ}{x^2}$ (point charge)
Two charges $+Q$ and $-Q$ are separated by distance $2a$. Find the electric field at a point on the perpendicular bisector at distance $r$ from midpoint.
Solution:
Distance from each charge: $\sqrt{r^2 + a^2}$
Field due to $+Q$: magnitude $E_1 = \frac{kQ}{r^2 + a^2}$, direction away from $+Q$
Field due to $-Q$: magnitude $E_2 = \frac{kQ}{r^2 + a^2}$, direction toward $-Q$
By symmetry: Components along the bisector cancel, perpendicular components add.
Component perpendicular to line joining charges:
$$E_{\perp} = 2E_1 \cos\theta = 2 \frac{kQ}{r^2 + a^2} \cdot \frac{a}{\sqrt{r^2 + a^2}}$$ $$\boxed{E = \frac{2kQa}{(r^2 + a^2)^{3/2}}}$$Direction: From $+Q$ to $-Q$ (perpendicular to bisector)
Note: This is the field of an electric dipole on equatorial line!
Quick Revision Box
| Concept | Formula/Key Point |
|---|---|
| Electric field definition | $\vec{E} = \frac{\vec{F}}{q_0}$ (force per unit charge) |
| Point charge field | $E = k\frac{Q}{r^2}$, direction: away from + / toward - |
| Superposition | $\vec{E}_{net} = \vec{E}_1 + \vec{E}_2 + ...$ (vector sum) |
| Field lines | Start from +, end at -, never cross, density = strength |
| Infinite line charge | $E = \frac{\lambda}{2\pi\varepsilon_0 r}$ |
| Infinite plane sheet | $E = \frac{\sigma}{2\varepsilon_0}$ (uniform!) |
| Spherical shell (inside) | $E = 0$ |
| Solid sphere (inside) | $E = k\frac{Qr}{R^3}$ (linear in $r$) |
| Conductor (electrostatic) | $E_{inside} = 0$, field perpendicular to surface |
JEE Weightage: 2-3 questions in JEE Main, 1-2 in JEE Advanced (often combined with other topics)
Time-Saver: For symmetric configurations, use symmetry to eliminate components before calculating!
Teacher’s Summary
- Electric field exists independently — it’s created by source charges, not by test charges
- Field is a vector — always use vector addition (superposition) for multiple sources
- Field lines never cross — unique field direction at each point
- Symmetry is your best friend — use it to simplify calculations dramatically
- Field inside conductor is zero — charges redistribute to cancel internal field
“The electric field is nature’s way of transmitting force. Master the field, and you master electrostatics.”
Related Topics
Within Electrostatics
- Electric Charges — The sources of electric field
- Electric Dipole — Field patterns for dipole
- Gauss’s Law — Advanced method for symmetric fields
- Electric Potential — Energy aspect of electric field
Connected Chapters
- Current Electricity — Electric field drives current
- Magnetism — Moving charges create magnetic fields
- Electromagnetic Waves — Oscillating electric and magnetic fields
Math Connections
- Vectors — Essential for field calculations
- Calculus — Integration for continuous distributions
- Coordinate Geometry — Field in 2D/3D space