Ever wondered why a balloon sticks to your hair after you rub it? That's electric field in action – an invisible force that nudges charges around.
In simple words, an electric field is a region where a charge feels a push or pull. Electric potential tells us how much energy a charge would have at a point, like how high a ball sits on a hill.
What is an Electric Field?
An electric field (symbol E) is a vector field – meaning it has both magnitude and direction – that tells us the force a tiny positive test charge would feel at any spot. Think of it like water flowing in a pipe: the speed of water tells you the field strength, and the direction the water moves tells you the field direction.
How to Find Electric Field Strength
The basic formula is E = F/q, where F is the force on a test charge q. For a point charge Q, Coulomb’s law gives E = k·Q / r², with k = 9×10⁹ N·m²/C² and r the distance from the charge.
What is Electric Potential?
Electric potential (symbol V) is the amount of electric potential energy per unit charge at a point. If you imagine a hill, the height of the hill is the potential – the higher you are, the more energy you have if you let go. The unit is the volt (V), equal to one joule per coulomb.
Potential Difference (Voltage)
Potential difference, often called voltage, is the change in potential between two points. It’s what drives current through a circuit, just like the slope of a hill makes a ball roll down.
Relationship Between Electric Field and Potential
The electric field is the spatial rate of change of potential. Mathematically, E = -dV/dx in one dimension, or 𝐄 = -∇V in three dimensions (∇ is the gradient operator). The negative sign means the field points from high to low potential, just as water flows downhill.
Worked Example: Point Charge
Suppose a point charge Q = 5 µC sits at the origin. Find the electric field and potential at a point 0.2 m away on the x‑axis.
- Step 1 – Potential: V = k·Q / r = (9×10⁹)(5×10⁻⁶) / 0.2 = 2.25×10⁵ V.
- Step 2 – Field: E = k·Q / r² = (9×10⁹)(5×10⁻⁶) / (0.2)² = 1.13×10⁶ N/C.
- Step 3 – Check Relation: Using E = -dV/dr ≈ V/r = 2.25×10⁵ V / 0.2 m = 1.13×10⁶ N/C, which matches the direct calculation (sign ignored for magnitude).
Quick Comparison: Electric Field vs. Electric Potential
| Aspect | Electric Field (E) | Electric Potential (V) |
|---|---|---|
| Nature | Vector (has direction) | Scalar (no direction) |
| Unit | Newton per coulomb (N/C) or volt per meter (V/m) | Volt (V) |
| What it tells you | Force on a unit positive charge | Energy per unit charge |
| Relation | E = -∇V | V = -∫E·dl (integral of field) |
| Analogy | Water flow speed and direction | Height of water surface |
Bullet Summary
- Electric field = push/pull on a test charge; visualise as arrows.
- Electric potential = stored energy per charge; think of height.
- Field points from high to low potential (negative gradient).
- For point charges, E ∝ 1/r², V ∝ 1/r.
- Use E = -dV/dx to move between the two concepts.
- Define electric field and give its SI unit.
Answer: Electric field is the force per unit positive charge at a point; unit N/C (or V/m). - State the relationship between electric field and electric potential.
Answer: E = -∇V; the field is the negative gradient of potential. - Calculate the electric field at 10 cm from a 2 µC point charge.
Answer: E = kQ/r² = (9×10⁹)(2×10⁻⁶)/(0.1)² = 1.8×10⁶ N/C. - A uniform electric field of 500 V/m exists between two parallel plates. What is the potential difference if the plates are 0.04 m apart?
Answer: V = E·d = 500 V/m × 0.04 m = 20 V. - Explain why electric field lines never cross each other.
Answer: Because at any point the field has a unique direction; crossing would imply two different directions simultaneously.