🌌 Introduction
The concept of an electric field is fundamental in physics and electrical engineering. It helps us understand how charged particles interact without touching each other. In this guide, we’ll break down the ideas of electric fields and electric field intensity in a clear and practical way.
🌟 What is an Electric Field?
An electric field is a region around a charged object where it can exert a force on other charged objects without any physical contact.
Definition:
The electric field is the space around a charged particle in which another charged particle experiences a force.
Symbol: \vec{E}
Unit: Newton per Coulomb (N/C) or Volt per meter (V/m)
🛠️ Real-Life Analogy
Imagine you’re placing a heater in the middle of a room. You don’t have to touch the heater to feel its warmth — the heat spreads through the air. Similarly, a charge creates an electric field in the space around it.
🔌 Electric Field Intensity (Strength)
Electric field intensity is the strength of the electric field at a specific point. It tells us how strong the force would be on a test charge placed at that point.
Formula: \vec{E} = \frac{\vec{F}}{q} Where:
- \vec{E} = Electric field intensity (N/C)
- \vec{F} = Force experienced by the charge (N)
- q = Test charge (C)
🧲 Electric Field Due to a Point Charge
Formula:
\boxed{E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{r^2}}
Where:
- Q = Source charge (C)
- r = Distance from the charge (m)
- \varepsilon_0 = Permittivity of free space = 8.85 \times 10^{-12} \ \text{C}^2/\text{N}\cdot\text{m}^2
⚖️ Units of Electric Field
- SI Unit: Newton/Coulomb (N/C)
- Alternative: Volt/meter (V/m)
🧭 Properties of Electric Field Lines
- Originate from positive and end on negative charges
- Never intersect
- Closer lines mean a stronger field
- Perpendicular to surface of conductor
🧮 Numerical Example
Q: Calculate the electric field at a distance of 0.2 m from a 5 \times 10^{-6} \ C charge.
Solution:
E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{r^2} = 9 \times 10^9 \cdot \frac{5 \times 10^{-6}}{(0.2)^2} = 1.125 \times 10^6 \ \text{N/C}
📈 Application in Daily Life
- Photocopiers
- CRT Monitors
- Lightning
- Electric dipoles in molecules
📚 Quick Summary
| Term | Symbol | Formula | Unit |
|---|---|---|---|
| Electric Field | \vec{E} | \vec{F}/q | N/C |
| Point Charge Field | EE | k \cdot Q/r^2 | N/C |
❓ Practice Questions
Q1: What is the electric field at a distance of 1 m from a 1μC= 1 \times 10^{-6} C charge?
Q2: If a charge of 2 \times 10^{-6} C feels a force of 0.1 N, what is the electric field intensity at that point?
✅ Answer Key with Explanation
Answer 1:
E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{r^2} = 9 \times 10^9 \cdot \frac{1 \times 10^{-6}}{1^2} = 9 \times 10^3 \ \text{N/C} = 9000 \ \text{N/C}
So, the electric field is 9000 N/C.
Answer 2: Using the formula:
E = \frac{0.1}{2 \times 10^{-6}} = 5 \times 10^4 = 50000 \ \text{N/C}
So, the electric field intensity is 50000 N/C.
