1. Concept Overview
A single point charge creates an electric potential around it that decreases as we move farther away. Unlike electric field, which is a vector, potential is a scalar quantity. This makes calculations much simpler because we only need to consider magnitude, not direction.
2. Definition
The electric potential at a distance \( r \) from a point charge \( Q \) is the work done per unit positive test charge in bringing it from infinity to that point.
The formula is:
\( V = \dfrac{1}{4\pi\varepsilon_0} \dfrac{Q}{r} \)
3. Understanding the Formula
The potential depends on:
- The magnitude of the charge \( Q \)
- The distance from the charge \( r \)
- The constant \( \dfrac{1}{4\pi\varepsilon_0} \)
3.1. Sign of Potential
- If \( Q > 0 \), potential is positive.
- If \( Q < 0 \), potential is negative.
This is because bringing a positive charge near a positive charge requires work, while bringing it near a negative charge releases energy.
3.2. Distance Dependence
The potential decreases with distance as \( 1/r \). This is slower than the \( 1/r^2 \) decrease seen in electric field.
4. Why Infinity Is Taken as the Reference
Potential is defined relative to some reference point. Infinity is chosen because as we go far away from a point charge, the electric field becomes extremely small. So the work done at infinity is effectively zero, making it a natural starting point.
5. Relation Between Electric Field and Potential
The electric field is related to the rate of change of potential.
5.1. Formula
For a point charge:
\( E = -\dfrac{dV}{dr} = \dfrac{1}{4\pi\varepsilon_0} \dfrac{Q}{r^2} \)
This shows how the potential formula and field formula are connected.
6. Superposition of Potentials
Since potential is a scalar, the total potential due to multiple point charges is simply the sum of individual potentials:
\( V_{\text{total}} = \sum_i \dfrac{1}{4\pi\varepsilon_0} \dfrac{Q_i}{r_i} \)
This makes calculations very easy when many charges are present.
7. Worked Examples
7.1. Example 1: Potential at a Point
A point charge \( Q = 4 \times 10^{-6} \text{ C} \) is placed at a distance \( r = 0.3 \text{ m} \). The potential is:
\( V = \dfrac{1}{4\pi\varepsilon_0} \dfrac{Q}{r} = 9 \times 10^9 \times \dfrac{4 \times 10^{-6}}{0.3} = 1.2 \times 10^5 \text{ V} \).
7.2. Example 2: Negative Charge
If \( Q = -5 \times 10^{-6} \text{ C} \) at a distance \( r = 0.5 \text{ m} \):
\( V = 9 \times 10^9 \times \dfrac{-5 \times 10^{-6}}{0.5} = -9 \times 10^4 \text{ V} \).
The negative sign shows that the potential decreases energy when bringing a positive test charge near.
8. Physical Interpretation
You can imagine electric potential as the electric "height" created by a point charge. A positive test charge naturally rolls downhill toward lower potential, just like a ball rolling down a slope. A positive point charge creates a high potential around it, and a negative charge creates a low potential. The closer you get to the charge, the stronger the effect.