Electrostatic Potential Due to a Single Charge

1. Concept Overview / Statement of the Law

Electrostatic Potential (also called Electric Potential) is a measure of how much electric potential energy a unit positive test charge possesses at a point in an electric field.

It answers a simple but fundamental question:

“How difficult or easy is it to bring a positive charge from infinity to a point in the electric field?”

To bring a charge into an electric field, we must do work against or with electric forces.
Thus:

Electrostatic Potential at a point is defined as the amount of work done per unit positive test charge in bringing the charge from infinity to that point, slowly, without acceleration.

Mathematically,

[
V = \dfrac{W}{q_0}
]

A point at higher potential is where a positive charge has more electric potential energy.
A point at lower potential is where the charge has less electric potential energy.

Important: Potential is a scalar quantity, unlike electric field which is a vector.

---

2. Explanation and Mathematical Derivation

Consider an electric field E created by some charge distribution.

We want to bring a small unit positive test charge [q_0] from infinity to a point P.

Work done by an external agent in moving the charge over a small displacement d𝑟 is:

[dW] [= -\vec{F} \cdot d\vec{r}] [= -q_0 \vec{E} \cdot d\vec{r}]

(The negative sign indicates that external work is done against electric force.)

 

Total work done in bringing the test charge from infinity to P:

[W = -q_0 \int_{\infty}^{P} \vec{E} \cdot d\vec{r}]

Electrostatic potential is defined as:

[
V(P) = \dfrac{W}{q_0}
]

So,

[
V(P) = -\int_{\infty}^{P} \vec{E} \cdot d\vec{r}
]

This formula directly links electric field and electric potential.

For a point charge [Q]:

[
E = \dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{r^2}
]

Using the above integral:

[
V = \dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{r}
]

---

3. Dimensions and Units

• Unit: Volt (V)
  [
  1 \text{ Volt} = 1 \dfrac{\text{Joule}}{\text{Coulomb}}
  ]
• Dimensions:
  [
  [V] = [ML^2T^{-3}A^{-1}]
  ]

---

4. Key Features

1. Potential is a scalar quantity.
2. Potential is defined relative to infinity (taken as zero potential).
3. It describes the energy per unit charge, not the force.
4. Potential is path-independent (electrostatic fields are conservative).
5. Work done in a closed loop is zero.
6. Higher potential means more energy for a positive charge.
7. Negative potential means the point is below reference (infinity) in energy.

---

5. Important Formulas (Table)

---

6. Conceptual Questions with Solutions (15)

---

7. FAQ / Common Misconceptions (10)

---

8. Practice Questions (With Step-by-Step Solutions)

---

Q1. What is the potential at a distance 20 cm from a point charge of +4 μC?

Solution:

[
V = \dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{r}
]

[
V = 9 \times 10^9 \times \dfrac{4\times10^{-6}}{0.2}
]

[
V = 1.8 \times 10^5 \text{ V}
]

---

Q2. At what distance from a +2 μC charge is the potential 5 × 10⁴ V?

[
V = \dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{r}
]

Rearranging:

[
r = \dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{V}
]

[
r = 9\times10^9 \times \dfrac{2\times10^{-6}}{5\times10^4}
]

[
r = 0.36 \text{ m}
]

---

Q3. The potential difference between two points is 20 V. How much work is needed to move a 4 C charge between them?

[
W = qV
]

[W] [= 4 \times 20] [= 80 \text{ J}]

---

Q4. A point is at zero potential. A +3 μC charge placed there has potential energy 0 J. What is the potential energy of –3 μC charge?

[U = qV] [= -3\mu C \times 0] [= 0]

Both have zero potential energy.

---

Q5. What is the potential at the center of a uniformly charged spherical shell of radius R?

Inside the shell, potential is constant:

[
V = \dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R}
]