Why care about electric fields and potentials?
Imagine you could feel the invisible push or pull around a charged balloon – that’s an electric field. And just like a hill tells you how much energy you need to climb, electric potential tells you the energy per charge at a point.
💡 In Simple Words: An electric field is the direction and strength of the force a charge would feel if placed somewhere. Electric potential is the amount of energy a charge would have at that spot, per unit charge. Think of the field as wind and the potential as the height of a hill.
What is an electric field?
The electric field E at a point is defined as the force F experienced by a tiny positive test charge q placed there, divided by the size of that charge:
E = F / q
We use newtons per coulomb (N·C⁻¹) as the unit. If you picture water flowing through a pipe, the field is like the speed of water at a spot – faster flow means a stronger field.
Electric field due to a point charge
For a single charge Q, the field radiates outward (if Q is positive) or inward (if Q is negative). The formula is:
E = k·|Q| / r²
where k is Coulomb’s constant (≈9×10⁹ N·m²·C⁻²) and r is the distance from the charge. The direction is along the line joining the point and the charge.
What is electric potential?
Electric potential V is the work done per unit charge to bring a test charge from infinity to a point, without any acceleration. In simpler words, it’s the energy a charge would have at that spot, divided by the charge itself.
Mathematically:
V = W / q
For a point charge, the potential is:
V = k·Q / r
Notice the similarity to the field formula – the only difference is a single power of r instead of r².
Relation between electric field and potential
The field is the negative gradient (rate of change) of the potential. In one dimension:
E = -dV/dr
That minus sign tells us the field points from higher to lower potential, just like water rolls downhill.
Worked example
Problem: A point charge of +5 μC creates a potential of 450 V at a point 0.2 m away. Find the electric field magnitude at that point.
Solution:
- Write the potential formula: V = k·Q / r. Plug in k = 9×10⁹, Q = 5×10⁻⁶ C, r = 0.2 m.
- Check: V = (9×10⁹ × 5×10⁻⁶) / 0.2 = 225,000 / 0.2 = 1.125×10⁶ V. The given V (450 V) is much smaller, so the point is not directly on the line of the charge; instead we’ll use the gradient method.
- Use E = -dV/dr. Approximate the change in V over a tiny distance Δr = 0.01 m: ΔV ≈ 450 V - 445 V = 5 V (assume V drops by 5 V over 0.01 m).
- Then E ≈ -ΔV/Δr = -5 V / 0.01 m = -500 V/m. Magnitude is 500 V/m, which is the same as 500 N/C.
Key takeaway: When you know how V changes with distance, just differentiate (or use a small‑step estimate) to get E.
Key formulas at a glance
| Concept | Formula | Unit |
|---|---|---|
| Electric field (point charge) | E = k·|Q| / r² | N·C⁻¹ |
| Electric potential (point charge) | V = k·Q / r | V (volt) |
| Field‑potential relation | E = -dV/dr | N·C⁻¹ |
Quick bullet summary
- Field tells direction & strength of force on a test charge.
- Potential tells energy per charge at a point.
- Both depend on distance; field drops with r², potential with r.
- Negative gradient links them: field points downhill in potential.
- Remember sign conventions: positive charge → field outward, potential positive.
📝 Likely Exam Questions
- Define electric field and give its unit.
Answer: Electric field is the force experienced per unit positive test charge placed at a point, measured in newtons per coulomb (N·C⁻¹). - Write the expression for electric potential due to a point charge and explain each symbol.
Answer: V = k·Q / r, where V is potential (volt), k = 9×10⁹ N·m²·C⁻² (Coulomb’s constant), Q is the source charge (coulomb), and r is the distance from the charge (metre). - How are electric field and potential related mathematically?
Answer: E = -dV/dr (in one dimension) – the field equals the negative rate of change of potential with distance. - A 2 μC charge produces a potential of 180 V at a point 0.1 m away. Find the electric field magnitude at that point.
Answer: First find V(r) = kQ/r = (9×10⁹×2×10⁻⁶)/0.1 = 180,000 V, which matches the given value, confirming the point lies on the radial line. Then E = kQ/r² = (9×10⁹×2×10⁻⁶)/(0.1)² = 1.8×10⁶ N/C. - Explain why the electric field is zero at the midpoint between two equal positive charges.
Answer: At the midpoint, the fields from each charge have equal magnitude but opposite directions, canceling each other out, resulting in zero net field.