Why electric fields matter in everyday life
Ever wondered why your hair stands up after you rub a balloon on your head? That tiny jump is a real‑world hint of an electric field at work.
An electric field is like an invisible wind that pushes charged objects. Electric potential is the height of a hill that tells you how much energy a charge would have at that spot.
What is an electric field?
The electric field (symbol E) is the force (push or pull) that a charge would feel per unit of test charge placed nearby. In plain words, it tells you how strong the “electric wind” is at any point.
We write it as E = F/q, where F is the force on a small test charge q. The unit is newton per coulomb (N C⁻¹).
Imagine water flowing through a pipe. The speed of the water at a point is like the electric field strength, and the tiny particle of water you drop is like the test charge.
How to calculate a field from a point charge
If you have a single charge Q (called a point charge) sitting alone, the field spreads out in all directions, just like ripples from a stone dropped in a pond. The formula is
E = k·Q / r²
Here k is Coulomb’s constant (≈9×10⁹ N m² C⁻²), and r is the distance from the charge to the point where you’re measuring.
Electric potential: the energy landscape
Electric potential (symbol V) is the amount of electric potential energy a unit charge would have at a point. Think of it as the height of a hill in a landscape of energy. Higher hill = more potential energy.
Potential is measured in volts (V), where 1 V = 1 joule per coulomb (J C⁻¹).
From charge to potential
For a single point charge, the potential at distance r is
V = k·Q / r
Notice the similarity to the field formula; the only difference is that the distance is not squared.
Relation between field and potential
The electric field points in the direction where the potential drops fastest, just like water rolls downhill. Mathematically, the field is the negative gradient (steepness) of the potential:
E = -dV/dr
In one‑dimensional problems you can simply take the derivative of V with respect to r and put a minus sign.
Worked example: point charge
Suppose a charge of +5 µC sits at the origin. Find the field and potential 10 cm away on the x‑axis.
- Convert units: 5 µC = 5×10⁻⁶ C, 10 cm = 0.10 m.
- Field: E = k·Q / r² = (9×10⁹)(5×10⁻⁶) / (0.10)² = (9×5×10³) / 0.01 = 4.5×10⁶ N C⁻¹. The direction is outward because the charge is positive.
- Potential: V = k·Q / r = (9×10⁹)(5×10⁻⁶) / 0.10 = 4.5×10⁵ V.
Notice the field is much larger than the potential because the distance appears squared in the denominator for the field.
Quick comparison: Electric field vs Electric potential
| Aspect | Electric Field (E) | Electric Potential (V) |
|---|---|---|
| What it tells you | Force per unit charge (strength of “electric wind”) | Energy per unit charge (height of “energy hill”) |
| Units | newton per coulomb (N C⁻¹) | volt = joule per coulomb (J C⁻¹) |
| Formula for point charge | E = kQ/r² | V = kQ/r |
| Relation | E = -dV/dr | Integral of –E gives V |
📝 Likely Exam Questions
- Define electric field and electric potential in your own words. Answer: Electric field is the force a unit positive test charge would feel at a point; electric potential is the potential energy per unit charge at that point.
- Derive the expression for the electric field of a point charge from its potential. Answer: Start with V = kQ/r, differentiate with respect to r, and add a minus sign: E = -dV/dr = kQ/r².
- A 2 µC charge creates a potential of 180 V at a point 0.05 m away. Find the electric field at that point. Answer: Use E = -dV/dr = kQ/r² = (9×10⁹)(2×10⁻⁶)/(0.05)² = 7.2×10⁶ N C⁻¹.
- Explain, with an analogy, why the electric field is stronger closer to a charge. Answer: Like water flowing faster near the opening of a hose, the “electric wind” is denser near the source, so the force per unit charge is larger.
- Sketch and label the electric field lines and equipotential lines around a positive point charge. Answer: Field lines radiate outward equally spaced; equipotentials are concentric circles centered on the charge.