Ever wondered why a balloon sticks to your hair after a static shock?
In simple words, an electric field is the invisible force that pushes or pulls other charges, while electric potential tells you how much energy a charge would have at a certain spot.
What is an Electric Field?
An electric field (we’ll call it E‑field) is a region around a charge where another charge feels a force. Think of it like the ripples that spread out when you drop a stone in a pond – the water moves outward, and anything floating on it gets nudged.
The direction of the field shows the direction a positive test charge would move. If the source charge is positive, the field points away; if it’s negative, the field points toward it.
How to Find the Electric Field of a Point Charge
For a single point charge Q, the field strength at a distance r is given by the formula:
E = k · Q / r²
Here k is Coulomb’s constant (≈9×10⁹ N·m²/C²). The farther you go, the weaker the field – just like the water ripples get smaller as they travel.
Example: A charge of +2 µC sits at the origin. What is the field 0.1 m away?
- Q = +2×10⁻⁶ C, r = 0.1 m
- E = (9×10⁹)·(2×10⁻⁶) / (0.1)² = 1.8×10⁶ N/C, directed radially outward.
What is Electric Potential?
Electric potential (often just called potential) is the amount of electric potential energy per unit charge at a point. If you imagine a hill, the height of the hill is like potential – the higher you are, the more energy you have to roll down.
Potential is measured in volts (V), where 1 V = 1 Joule per Coulomb (J/C). A positive charge placed at a high‑potential spot wants to move to lower potential, just as a ball rolls downhill.
Potential Difference (Voltage) Explained
The term potential difference or voltage is the difference in potential between two points. It’s what drives current through a circuit, similar to how a slope drives water to flow.
For a point charge Q, the potential at distance r is:
V = k·Q / r
Notice the r is not squared – potential drops slower with distance than the field does.
Relationship Between Electric Field and Potential
The field and potential are two sides of the same coin. Mathematically, the electric field is the negative gradient (rate of change) of the potential:
E = - dV/dr
In plain English: if the potential drops quickly over a short distance, the field is strong. The minus sign just tells us the field points from high to low potential.
Worked Example: From Potential to Field
Suppose the potential around a charge varies as V(r) = 5 × 10⁴ / r (with r in meters). Find the field at r = 0.2 m.
- Differentiate V: dV/dr = -5×10⁴ / r²
- Apply E = -dV/dr → E = 5×10⁴ / r²
- Plug r = 0.2 m: E = 5×10⁴ / (0.2)² = 5×10⁴ / 0.04 = 1.25×10⁶ N/C, directed outward.
Key Differences at a Glance
| Aspect | Electric Field (E) | Electric Potential (V) |
|---|---|---|
| What it measures | Force per unit positive charge (N/C) | Energy per unit charge (J/C or V) |
| Units | Newtons per Coulomb (N/C) or Volts per meter (V/m) | Volts (V) |
| Direction | Vector (has direction) | Scalar (no direction) |
| Formula for a point charge | E = kQ/r² | V = kQ/r |
| How it changes with distance | Falls as 1/r² | Falls as 1/r |
📝 Likely Exam Questions
- Q1. Define electric field and state its unit.
Answer: Electric field is the force experienced per unit positive test charge placed in the region around a source charge. Unit: newton per coulomb (N/C) or volt per metre (V/m). - Q2. A point charge of +3 µC creates a potential of 540 V at a point. Find the distance of that point from the charge.
Answer: Use V = kQ/r → r = kQ/V = (9×10⁹)(3×10⁻⁶)/540 ≈ 0.05 m. - Q3. Explain why the electric field is zero at the midpoint between two equal positive charges.
Answer: The fields due to each charge have equal magnitude but opposite directions at the midpoint, cancelling each other out. - Q4. Derive the relationship E = -dV/dr for a radial field.
Answer: Starting from V(r) = kQ/r, differentiate: dV/dr = -kQ/r². The field magnitude is E = kQ/r², so E = -dV/dr, pointing from high to low potential. - Q5. A uniform electric field of 200 N/C exists between two parallel plates. What is the potential difference if the plates are 0.04 m apart?
Answer: V = E·d = 200 N/C × 0.04 m = 8 V.