Why electric fields matter in everyday life
Ever felt a static shock after rubbing a sweater? That's an electric field at work, nudging electrons from one place to another.
💡 In Simple Words: An electric field is like an invisible wind that pushes tiny charged particles. Electric potential is the “height” of that wind, telling you how much energy a charge would gain if it moves.
Electric field – the invisible wind
The electric field (symbol E) is the force per unit positive charge. In plain words, if you place a tiny +1 C charge somewhere, the force it feels divided by 1 C gives you the field there.
- Definition: E = F/q, where F is the force on a test charge q.
- Direction: points away from positive charges and toward negative ones, just like water flows away from a source.
- Units: newtons per coulomb (N C⁻¹), which is the same as volts per meter (V m⁻¹).
How to calculate the field of a point charge
Imagine a single charge Q sitting at the centre of a sphere. The field at a distance r is
E = k Q / r²
Here k is Coulomb’s constant (≈9×10⁹ N m² C⁻²). The formula looks just like the gravity law, only with charge instead of mass.
Electric potential – the energy “height”
Electric potential (symbol V) tells you how much electric potential energy a unit charge would have at a point.
- Definition: V = U/q, where U is the potential energy of a charge q.
- It’s measured in volts (V), where 1 V = 1 joule per coulomb (J C⁻¹).
- Think of a hill: the higher you are, the more potential energy you have. Electric potential works the same way for charges.
Potential of a point charge
For a single charge Q, the potential at distance r is
V = k Q / r
Notice the similarity to the field formula – the field drops with r² while potential drops with r.
Relationship between electric field and potential
The field is the slope of the potential landscape. Mathematically, the field is the negative gradient (rate of change) of potential:
E = –∇V
In one dimension this simply becomes E = –dV/dx. So if the potential changes quickly over a short distance, the field is strong.
Worked example: field from a known potential
Suppose the potential along a line varies as V(x) = 5 – 2x (V, with x in metres). Find the electric field at x = 1 m.
- Take the derivative: dV/dx = –2 V m⁻¹.
- Insert the negative sign: E = –( –2 ) = +2 V m⁻¹, i.e., 2 N C⁻¹ directed in the +x direction.
Quick comparison
| Aspect | Electric Field (E) | Electric Potential (V) |
|---|---|---|
| What it tells you | Force on a unit positive charge | Energy per unit charge |
| Units | N C⁻¹ or V m⁻¹ | V (J C⁻¹) |
| Direction | Vector (has direction) | Scalar (no direction) |
| Formula for point charge | E = kQ/r² | V = kQ/r |
| Relation | E = –∇V | V = –∫E·dl |
Tips for exam problems
- Identify whether the question asks for a vector (field) or a scalar (potential).
- Remember the sign convention: field points from high to low potential.
- When charges are multiple, use superposition – add fields (vectors) and potentials (scalars) separately.
- For uniform fields, E = V/d, where d is the separation between two equipotential surfaces.
📝 Likely Exam Questions
- Define electric field and state its unit.
Answer: Electric field is the force experienced per unit positive charge, measured in newtons per coulomb (N C⁻¹) or volts per metre. - Two point charges +3 μC and –3 μC are 0.2 m apart. Find the magnitude of the electric field at the midpoint.
Answer: Each charge produces E = kQ/r² with r = 0.1 m. E = (9×10⁹)(3×10⁻⁶)/(0.1)² = 2.7×10⁶ N C⁻¹. The fields point in opposite directions, so they add to 5.4×10⁶ N C⁻¹. - What is the potential difference between two points 4 cm apart in a uniform field of 500 V m⁻¹?
Answer: ΔV = E·d = 500 V m⁻¹ × 0.04 m = 20 V. - Explain why electric field lines never cross.
Answer: If they crossed, a point would have two different field directions, which contradicts the definition of a unique vector field at each location. - Derive the relationship E = –dV/dx for a one‑dimensional field.
Answer: Starting from V = –∫E dx, differentiate both sides with respect to x to get dV/dx = –E, hence E = –dV/dx.