Electric Potential: Complete Guide to Potential Difference, Point Charges & Equipotential Surfaces | HRK Physics

Chapter 30: Electric Potential

Complete Physics Notes Based on Halliday, Resnick and Krane - B.Sc. Physics Edition 2015-16

Electric Potential Potential Difference Electromagnetism Reading Time: 25 min

📋 Table of Contents

• 1. Introduction to Electric Potential
• 2. Potential Difference
• 3. Absolute Electric Potential
• 4. Potential Difference due to Point Charge
• 5. Absolute Potential due to Point Charge
• 6. Electric Potential for Collection of Charges
• 7. Electric Potential due to a Dipole
• 8. Electric Potential of Continuous Charge Distribution
• 9. Equipotential Surfaces
• 10. Calculating Electric Field from Potential
• 11. Electric Potential and Field of Conductors
• 12. Solved Problems
• Frequently Asked Questions

Introduction to Electric Potential

⚡ Energy Approach in Electrostatics

The energy approach in the study of dynamics of particles can yield not only simplification but also new insights. One advantage of the energy method is that, although force is a vector, energy is a scalar. In problems involving vector forces and fields, calculations involving sums and integrals are often complicated. In this chapter, we introduce the energy method to the study of electrostatics.

🔬 Scalar vs Vector Approach

While electric field is a vector quantity that requires complex vector calculations, electric potential is a scalar quantity that simplifies many electrostatic problems. This scalar approach provides deeper insights into the behavior of electric fields and charged particles.

Potential Difference

📏 Definition of Potential Difference

Potential difference '\(\Delta V\)' between two points is defined as "the amount of work done '\(\Delta W\)' per unit charge '\(q_0\)' in moving it from one point to the other against the electric field and by keeping the system in equilibrium".

\[ \Delta V = \frac{\Delta W}{q_0} \]

Work Done Against Electric Field

Suppose a unit positive test charge '\(q_0\)' is moved from point '\(a\)' to point '\(b\)' in the electric field '\(\vec{E}\)' of a large positive charge '\(q\)'.

The work done in moving '\(q_0\)' from point '\(a\)' to point '\(b\)' against the electric field '\(\vec{E}\)' is:

\[ W_{a \rightarrow b} = \int_a^b \vec{F} \cdot \vec{dr} \]

Force Required

The electrical force of magnitude '\(\vec{F} = -q_0 \vec{E}\)' must be supplied to move '\(q_0\)' against the electric field:

\[ W_{a \rightarrow b} = \int_a^b (-q_0 \vec{E}) \cdot \vec{dr} \]

\[ = -q_0 \int_a^b \vec{E} \cdot \vec{dr} \]

Potential Difference Formula

Dividing both sides by \(q_0\):

\[ \frac{W_{a \rightarrow b}}{q_0} = -\int_a^b \vec{E} \cdot \vec{dr} \]

Since \(\frac{W_{a \rightarrow b}}{q_0} = \Delta V = V_b - V_a\), we get:

\[ V_b - V_a = -\int_a^b \vec{E} \cdot \vec{dr} \]

Absolute Electric Potential

🎯 Definition of Absolute Electric Potential

Absolute electric potential at a point is defined as "the amount of work done per unit charge in moving it from infinity to a specific field point against the electric field and by keeping the system in equilibrium".

Reference Point at Infinity

To find the absolute potential, the reference point is selected where potential is zero. This point is situated at infinity (out of the electric field). Thus:

\[ V_a = V(\infty) = 0 \]

General Formula

From the potential difference formula:

\[ V_b - 0 = -\int_\infty^b \vec{E} \cdot \vec{dr} \]

If the distance from point '\(b\)' to the charge '\(q\)' is '\(r\)', then in general:

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

Potential Difference due to Point Charge

🔋 Setup for Point Charge

The potential difference between two points is the amount of work done per unit charge '\(q_0\)' in moving it from one point to the other against the electric field '\(E\)'.

Start with General Formula

\[ V_b - V_a = -\int_{r_a}^{r_b} \vec{E} \cdot \vec{dr} \]

Electric Field of Point Charge

The electric field intensity due to a point charge:

\[ \vec{E} = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r} \]

Substituting into the potential difference formula:

\[ V_b - V_a = -\int_{r_a}^{r_b} \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{r} \cdot \vec{dr} \]

\[ = -\frac{q}{4\pi\epsilon_0} \int_{r_a}^{r_b} \frac{\hat{r} \cdot \vec{dr}}{r^2} \]

Dot Product Calculation

Since \(\vec{E}\) is directed radially outward, \(\hat{r} \cdot \vec{dr} = |\hat{r}| |\vec{dr}| \cos 0^\circ = (1)(dr)(1) = dr\):

\[ V_b - V_a = -\frac{q}{4\pi\epsilon_0} \int_{r_a}^{r_b} \frac{dr}{r^2} \]

Integration

\[ V_b - V_a = -\frac{q}{4\pi\epsilon_0} \left| -\frac{1}{r} \right|_{r_a}^{r_b} \]

\[ = \frac{q}{4\pi\epsilon_0} \left| \frac{1}{r} \right|_{r_a}^{r_b} \]

\[ = \frac{q}{4\pi\epsilon_0} \left[ \frac{1}{r_b} - \frac{1}{r_a} \right] \]

📐 Final Expression

The potential difference between two points 'a' and 'b' due to a point charge is:

\[ V_b - V_a = \frac{q}{4\pi\epsilon_0} \left[ \frac{1}{r_b} - \frac{1}{r_a} \right] \]

Absolute Potential due to Point Charge

🎯 Electric Potential at a Point

The electric potential at any point is the amount of work done per unit charge in moving a unit positive charge (test charge) from infinity to that point, against the electric field.

Set Reference at Infinity

If point 'a' is at infinity:

\[ V_a = V(\infty) = 0, \quad r_a = \infty \]

Substitute into Potential Difference Formula

\[ V_b - 0 = \frac{q}{4\pi\epsilon_0} \left[ \frac{1}{r_b} - \frac{1}{\infty} \right] \]

Since \(\frac{1}{\infty} = 0\):

\[ V_b = \frac{q}{4\pi\epsilon_0} \left[ \frac{1}{r_b} \right] \]

General Formula

In general, the electric potential at a point due to a point charge 'q' is:

\[ V = \frac{1}{4\pi\epsilon_0} \frac{q}{r} \]

Sample Problem 1: Potential Energy Between Protons

Question: Two protons in the nucleus of \( U^{238} \) are 6 fm apart. What is the potential energy associated with the electric force that acts between them?

Solution:

Given: \( r = 6 \, \text{fm} = 6 \times 10^{-15} \, \text{m} \)

\( q_1 = q_2 = 1e = 1.6 \times 10^{-19} \, \text{C} \)

\[ \Delta V = k \frac{q}{r} = 9 \times 10^9 \times \frac{1.6 \times 10^{-19}}{6 \times 10^{-15}} \]

\[ = 2.34 \times 10^5 \, \text{V} \]

\[ \Delta U = q_1 \cdot \Delta V = 1.6 \times 10^{-19} \times 2.34 \times 10^5 \]

\[ = 3.744 \times 10^{-14} \, \text{J} \]

\[ \Rightarrow \Delta U = \frac{3.744 \times 10^{-14}}{1.6 \times 10^{-19}} \, \text{eV} \]

\[ = 2.34 \times 10^5 \, \text{eV} \]

Sample Problem 5: Finding Charge from Potential

Question: What must be the magnitude of an isolated positive point charge for the electric potential at 15 cm from the charge to be 120 V?

Solution:

Given: \( r = 15 \, \text{cm} = 15 \times 10^{-2} \, \text{m} \), \( V = 120 \, \text{V} \)

Using \( V = k \frac{q}{r} \):

\[ q = \frac{V \cdot r}{k} = \frac{120 \times 15 \times 10^{-2}}{9 \times 10^9} \]

\[ = 1.995 \times 10^{-9} \, \text{C} \]

Electric Potential for Collection of Charges

🔢 Superposition Principle for Potential

Let "\( q_1, q_2, q_3, \ldots, q_n \)" be 'n' point charges at distances "\( r_1, r_2, r_3, \ldots, r_n \)" from a point 'P'. The total electric potential at point 'P' due to this collection of charges is the scalar sum of the potentials due to individual charges.

Individual Potentials

\[ V_1 = \frac{1}{4\pi\epsilon_0} \frac{q_1}{r_1} \]

\[ V_2 = \frac{1}{4\pi\epsilon_0} \frac{q_2}{r_2} \]

\[ V_3 = \frac{1}{4\pi\epsilon_0} \frac{q_3}{r_3} \]

\[ \vdots \]

\[ V_n = \frac{1}{4\pi\epsilon_0} \frac{q_n}{r_n} \]

Total Potential

\[ V = V_1 + V_2 + V_3 + \ldots + V_n \]

\[ = \frac{1}{4\pi\epsilon_0} \frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0} \frac{q_2}{r_2} + \frac{1}{4\pi\epsilon_0} \frac{q_3}{r_3} + \ldots + \frac{1}{4\pi\epsilon_0} \frac{q_n}{r_n} \]

\[ = \frac{1}{4\pi\epsilon_0} \left( \frac{q_1}{r_1} + \frac{q_2}{r_2} + \frac{q_3}{r_3} + \ldots + \frac{q_n}{r_n} \right) \]

\[ = \frac{1}{4\pi\epsilon_0} \sum_{i=1}^n \frac{q_i}{r_i} \]

Sample Problem 9: Potential of Two Charges

Question: Two charges \( q_1 = +12 \, \text{nC} \) and \( q_2 = -12 \, \text{nC} \) are placed 10 cm apart. Compute the potentials at points a, b, and c.

Solution:

At point a:

\[ V_a = k \left( \frac{q_1}{r_1} + \frac{q_2}{r_2} \right) \]

\[ = 9 \times 10^9 \left( \frac{12 \times 10^{-9}}{0.06} + \frac{-12 \times 10^{-9}}{0.04} \right) \]

\[ = 9 \times 10^9 (200 \times 10^{-9} - 300 \times 10^{-9}) \]

\[ = -900 \, \text{V} \]

At point b:

\[ V_b = k \left( \frac{q_1}{r_1} + \frac{q_2}{r_2} \right) \]

\[ = 9 \times 10^9 \left( \frac{12 \times 10^{-9}}{0.04} + \frac{-12 \times 10^{-9}}{0.14} \right) \]

\[ = 9 \times 10^9 (300 \times 10^{-9} - 85.7 \times 10^{-9}) \]

\[ = 1928.7 \, \text{V} \]

At point c:

\[ V_c = k \left( \frac{q_1}{r_1} + \frac{q_2}{r_2} \right) \]

\[ = 9 \times 10^9 \left( \frac{12 \times 10^{-9}}{0.16} + \frac{-12 \times 10^{-9}}{0.16} \right) \]

\[ = 0 \, \text{V} \]

Electric Potential due to a Dipole

🔋 Electric Dipole Setup

An electric dipole consists of two charges of equal magnitude but opposite sign separated by a distance '2a'. The dipole moment '\( \vec{p} \)' is defined as \( \vec{p} = q \cdot 2\vec{a} \).

Potential at Point P

Let P be a point at distance 'r' from the center of the dipole, making an angle 'θ' with the dipole axis.

\[ V = V_+ + V_- = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{r_+} - \frac{q}{r_-} \right) \]

Approximation for Large r

For \( r \gg a \), we can use the approximations:

\[ r_+ \approx r - a \cos\theta \]

\[ r_- \approx r + a \cos\theta \]

\[ \frac{1}{r_+} - \frac{1}{r_-} \approx \frac{2a \cos\theta}{r^2} \]

Final Expression

\[ V = \frac{1}{4\pi\epsilon_0} \frac{q \cdot 2a \cos\theta}{r^2} \]

\[ = \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2} \]

\[ = \frac{1}{4\pi\epsilon_0} \frac{\vec{p} \cdot \hat{r}}{r^2} \]

📊 Special Cases

• Along the axis (θ = 0°): \( V = \frac{1}{4\pi\epsilon_0} \frac{p}{r^2} \)
• Perpendicular to axis (θ = 90°): \( V = 0 \)
• Opposite to axis (θ = 180°): \( V = -\frac{1}{4\pi\epsilon_0} \frac{p}{r^2} \)

Electric Potential of Continuous Charge Distribution

📐 General Approach

For a continuous charge distribution, we divide the distribution into infinitesimal charge elements 'dq'. The potential at point P due to the entire distribution is the sum of potentials due to all charge elements.

Potential due to Charge Element

\[ dV = \frac{1}{4\pi\epsilon_0} \frac{dq}{r} \]

where 'r' is the distance from the charge element to point P.

Total Potential

\[ V = \int dV = \frac{1}{4\pi\epsilon_0} \int \frac{dq}{r} \]

Line Charge Distribution

For a line charge with linear charge density λ:

\[ V = \frac{1}{4\pi\epsilon_0} \int \frac{\lambda dl}{r} \]

Surface Charge Distribution

For a surface charge with surface charge density σ:

\[ V = \frac{1}{4\pi\epsilon_0} \int \frac{\sigma dA}{r} \]

Volume Charge Distribution

For a volume charge with volume charge density ρ:

\[ V = \frac{1}{4\pi\epsilon_0} \int \frac{\rho dV}{r} \]

Sample Problem 15: Potential of Charged Ring

Question: A thin plastic rod bent into a ring of radius R has a uniformly distributed charge Q. What is the electric potential at point P on the axis of the ring at distance z from the center?

Solution:

Linear charge density: \( \lambda = \frac{Q}{2\pi R} \)

\[ dV = \frac{1}{4\pi\epsilon_0} \frac{dq}{r} = \frac{1}{4\pi\epsilon_0} \frac{\lambda dl}{\sqrt{R^2 + z^2}} \]

\[ V = \int dV = \frac{1}{4\pi\epsilon_0} \frac{\lambda}{\sqrt{R^2 + z^2}} \int_0^{2\pi R} dl \]

\[ = \frac{1}{4\pi\epsilon_0} \frac{\lambda}{\sqrt{R^2 + z^2}} \cdot 2\pi R \]

\[ = \frac{1}{4\pi\epsilon_0} \frac{Q}{\sqrt{R^2 + z^2}} \]

Equipotential Surfaces

🎯 Definition

An equipotential surface is a surface on which the electric potential is the same at every point. No work is required to move a charge from one point to another on an equipotential surface.

Point Charge

Equipotential surfaces are concentric spheres centered on the charge.

Uniform Electric Field

Equipotential surfaces are planes perpendicular to the field lines.

Electric Dipole

Equipotential surfaces are more complex, with potential decreasing with distance from the dipole.

🔍 Properties of Equipotential Surfaces

• Electric field lines are always perpendicular to equipotential surfaces.
• No work is done when moving a charge along an equipotential surface.
• Equipotential surfaces never cross each other.
• The spacing between equipotential surfaces indicates the strength of the electric field.

Calculating Electric Field from Potential

📐 Potential Gradient

The electric field is related to the potential by the gradient operator. In one dimension:

\[ E_x = -\frac{dV}{dx} \]

General Relationship

In three dimensions, the electric field is the negative gradient of the potential:

\[ \vec{E} = -\nabla V \]

\[ = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right) \]

Component Form

\[ E_x = -\frac{\partial V}{\partial x} \]

\[ E_y = -\frac{\partial V}{\partial y} \]

\[ E_z = -\frac{\partial V}{\partial z} \]

Sample Problem 21: Field from Potential

Question: The electric potential in a region of space is given by \( V = 3x^2 - 2y^2 + 4z \). Find the electric field at point (2, 3, 5).

Solution:

Using \( \vec{E} = -\nabla V \):

\[ E_x = -\frac{\partial V}{\partial x} = -\frac{\partial}{\partial x}(3x^2 - 2y^2 + 4z) \]

\[ = -6x \]

\[ E_y = -\frac{\partial V}{\partial y} = -\frac{\partial}{\partial y}(3x^2 - 2y^2 + 4z) \]

\[ = 4y \]

\[ E_z = -\frac{\partial V}{\partial z} = -\frac{\partial}{\partial z}(3x^2 - 2y^2 + 4z) \]

\[ = -4 \]

At point (2, 3, 5):

\[ E_x = -6(2) = -12 \, \text{V/m} \]

\[ E_y = 4(3) = 12 \, \text{V/m} \]

\[ E_z = -4 \, \text{V/m} \]

\[ \vec{E} = (-12, 12, -4) \, \text{V/m} \]

Electric Potential and Field of Conductors

⚡ Properties of Conductors

For conductors in electrostatic equilibrium:

Property 1

The electric field inside a conductor is zero.

\[ E_{\text{inside}} = 0 \]

Property 2

The entire conductor is at the same potential.

\[ V_{\text{conductor}} = \text{constant} \]

Property 3

The electric field just outside a conductor is perpendicular to the surface.

🔍 Implications

• No work is required to move a charge from one point to another on a conductor.
• The surface of a conductor is an equipotential surface.
• Charge resides only on the surface of a conductor.

Solved Problems

Problem 25: Potential Energy of Three Charges

Question: Three charges \( q_1 = +1 \mu C \), \( q_2 = -2 \mu C \), and \( q_3 = +3 \mu C \) are placed at the corners of an equilateral triangle of side 1 m. Calculate the total potential energy of the system.

Solution:

Potential energy of a system of charges:

\[ U = \frac{1}{4\pi\epsilon_0} \left( \frac{q_1 q_2}{r_{12}} + \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}} \right) \]

\[ = 9 \times 10^9 \left( \frac{(1 \times 10^{-6})(-2 \times 10^{-6})}{1} + \frac{(1 \times 10^{-6})(3 \times 10^{-6})}{1} + \frac{(-2 \times 10^{-6})(3 \times 10^{-6})}{1} \right) \]

\[ = 9 \times 10^9 (-2 \times 10^{-12} + 3 \times 10^{-12} - 6 \times 10^{-12}) \]

\[ = 9 \times 10^9 (-5 \times 10^{-12}) \]

\[ = -4.5 \times 10^{-2} \, \text{J} \]

Problem 30: Potential of Charged Disk

Question: A disk of radius R has a uniform surface charge density σ. Find the electric potential at a point on the axis of the disk at distance z from the center.

Solution:

Divide the disk into rings of radius r and width dr:

\[ dq = \sigma \cdot 2\pi r dr \]

\[ dV = \frac{1}{4\pi\epsilon_0} \frac{dq}{\sqrt{r^2 + z^2}} = \frac{1}{4\pi\epsilon_0} \frac{\sigma \cdot 2\pi r dr}{\sqrt{r^2 + z^2}} \]

\[ V = \int_0^R dV = \frac{\sigma}{2\epsilon_0} \int_0^R \frac{r dr}{\sqrt{r^2 + z^2}} \]

Let \( u = r^2 + z^2 \), \( du = 2r dr \):

\[ V = \frac{\sigma}{2\epsilon_0} \cdot \frac{1}{2} \int_{z^2}^{R^2 + z^2} \frac{du}{\sqrt{u}} \]

\[ = \frac{\sigma}{4\epsilon_0} \left[ 2\sqrt{u} \right]_{z^2}^{R^2 + z^2} \]

\[ = \frac{\sigma}{2\epsilon_0} \left( \sqrt{R^2 + z^2} - z \right) \]

Frequently Asked Questions

❓ What is the difference between electric potential and electric potential energy?

Electric potential is the potential energy per unit charge at a point in an electric field, while electric potential energy is the energy a charge has due to its position in an electric field. Electric potential is a property of the field itself, while electric potential energy depends on both the field and the charge placed in it.

❓ Why is electric potential a scalar quantity while electric field is a vector?

Electric potential is defined as work done per unit charge, and work is a scalar quantity. The electric field, on the other hand, represents the force per unit charge, and force is a vector quantity. This scalar nature of potential makes calculations simpler in many cases.

❓ Why is the potential inside a conductor constant?

In electrostatic equilibrium, the electric field inside a conductor is zero. Since the electric field is the negative gradient of potential (\( \vec{E} = -\nabla V \)), a zero field implies that the potential doesn't change with position inside the conductor, making it constant throughout.

❓ Can electric potential be negative?

Yes, electric potential can be negative. The sign of potential depends on the reference point (usually taken as infinity where potential is zero) and the sign of the charges creating the potential. Negative potential means that work would need to be done to bring a positive test charge from infinity to that point.

📚 Continue Your Physics Journey

Mastering Electric Potential is fundamental to understanding electromagnetism. These comprehensive notes based on Halliday, Resnick and Krane provide a solid foundation for further studies in physics including capacitance and current electricity.

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© 2025 Physics Education Initiative | Chapter 30: Electric Potential

These comprehensive notes are designed to help B.Sc. Physics students understand fundamental concepts of Electric Potential based on Halliday, Resnick and Krane

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