Why should you care about electric fields?
Imagine you could feel the invisible push or pull around a charged balloon – that’s an electric field in action, shaping everything from lightning to your phone’s touchscreen.
💡 In Simple Words: An electric field is the space around a charge where another charge would feel a force. Electric potential tells us how much energy a charge would have at a point, like the height of a hill that a rolling ball would gain or lose.
What is an Electric Field?
Electric field (E) is a vector quantity – it has both size and direction. It tells us the force a tiny positive test charge would feel if placed at that spot, divided by the amount of charge on the test charge.
Mathematically, E = F/q, where F is the force and q is the test charge. Think of water flowing through a pipe: the pressure difference drives the flow, just as a charge creates a “pressure” that pushes other charges.
How to picture it
- Draw arrows (field lines) pointing away from a positive charge and toward a negative charge.
- Denser lines mean a stronger field, like more traffic lanes on a busy road.
What is Electric Potential?
Electric potential (V) is the amount of electric potential energy per unit charge at a point. It’s a scalar – only a size, no direction.
Formula: V = U/q, where U is the potential energy. If you imagine a hill, the height of the hill is the potential. A ball (charge) placed higher up has more energy ready to roll down.
Potential difference (Voltage)
The difference in potential between two points, called voltage, tells us how much energy a charge will gain or lose when moving between them. It’s what makes current flow in a circuit.
Relationship Between Field and Potential
The electric field is the spatial rate of change of potential. In simple terms, if you move a little distance Δs in the direction of the field, the potential drops by E·Δs.
Mathematically, E = -∇V (the negative gradient of V). The minus sign means the field points from high potential to low potential – just like water flows downhill.
Worked Example
Problem: A point charge +5 µC creates a field at a point 0.10 m away. Find the electric field magnitude and the potential at that point. (Use k = 9×10⁹ N·m²/C².)
- Electric field: E = k·q/r² → E = 9×10⁹·5×10⁻⁶ / (0.10)² = 4.5×10⁶ N/C.
- Potential: V = k·q/r → V = 9×10⁹·5×10⁻⁶ / 0.10 = 4.5×10⁵ V.
Notice the field is huge because it drops with the square of distance, while potential falls off linearly.
Quick Comparison – Electric Field vs. Electric Potential
| Aspect | Electric Field (E) | Electric Potential (V) |
|---|---|---|
| Type | Vector (has direction) | Scalar (no direction) |
| Unit | Newton per coulomb (N/C) or volt per meter (V/m) | Volt (V) |
| Physical meaning | Force on a unit positive charge | Energy per unit charge |
| How it changes with distance | ∝ 1/r² for a point charge | ∝ 1/r for a point charge |
| Relation | E = -∇V (gradient of V) | V = -∫E·ds (integral of E) |
How to Solve Typical ISC Problems
Follow this simple checklist whenever a question asks for field or potential.
Key Points to Remember
- Field lines start on positive charges and end on negative charges.
- Closer field lines → stronger field.
- Potential is highest near positive charges and lowest near negatives.
- The field points downhill on the potential “hill”.
- Use E = kq/r² for field and V = kq/r for potential of isolated point charges.
- When charges are many, add contributions (superposition).
📝 Likely Exam Questions
- Define electric field and state its SI unit.
Answer: Electric field is the force experienced per unit positive test charge placed at a point. Unit: newton per coulomb (N/C) or volt per meter (V/m). - Calculate the electric potential at a point 5 cm from a charge of –2 µC.
Answer: V = k·q/r = 9×10⁹·(‑2×10⁻⁶)/0.05 = ‑3.6×10⁵ V. - Two equal positive charges are 0.20 m apart. Find the magnitude of the electric field at the midpoint.
Answer: Each charge creates E = kq/(0.10)² = 9×10⁹·(1×10⁻⁶)/0.01 = 9×10⁸ N/C. Directions are opposite, so they cancel; net field = 0. - Explain why electric field lines are never closed loops.
Answer: Field lines start on positive charges and end on negative charges because electric field is a gradient of potential; there’s no source or sink in free space, so they cannot loop back on themselves. - State the relationship between electric field and potential difference.
Answer: The electric field is the negative gradient of potential, E = –∇V, meaning the field points from higher to lower potential.