Electric Charge and Coulomb's Law

Electric Charge

Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electric and magnetic field. There are two types of electric charges: positive and negative. Like charges repel each other, while unlike charges attract.

Properties of Electric Charge

Addition of Charges: The total charge in a system of point charges is the algebraic sum of individual charges.

Conservation of Charge: The total charge in an isolated system remains constant; charge can neither be created nor destroyed.

Quantisation of Charge: Electric charge exists in discrete amounts, integral multiples of the elementary charge \( e = 1.6 \times 10^{-19} \) coulombs.

Coulomb's Law

Coulomb's law quantifies the electrostatic force between two point charges. The magnitude of the force \( F \) between charges \( q_1 \) and \( q_2 \) separated by distance \( r \) is given by:

\[ F = k \frac{|q_1 q_2|}{r^2} \]

where \( k = \frac{1}{4 \pi \varepsilon_0} = 8.99 \times 10^9 \ \text{N m}^2/\text{C}^2 \) is Coulomb's constant, and \( \varepsilon_0 = 8.854 \times 10^{-12} \ \text{F/m} \) is the permittivity of free space.

The force acts along the line joining the charges and is attractive if charges are opposite, repulsive if charges are alike.

Vector Form of Coulomb's Law

The force on charge \( q_1 \) due to \( q_2 \) is:

\[ \vec{F}_{12} = k \frac{q_1 q_2}{r^2} \hat{r}_{12} \]

where \( \hat{r}_{12} \) is the unit vector from \( q_1 \) to \( q_2 \). Newton's third law implies \( \vec{F}_{21} = -\vec{F}_{12} \).

Solved Examples

Example 1: Calculate the force between two charges of +3 \( \mu \)C and -2 \( \mu \)C placed 0.5 m apart in vacuum.

Solution:

Given: \( q_1 = 3 \times 10^{-6} \) C, \( q_2 = -2 \times 10^{-6} \) C, \( r = 0.5 \) m

Using Coulomb's law:

\[ F = k \frac{|q_1 q_2|}{r^2} = 8.99 \times 10^9 \times \frac{(3 \times 10^{-6})(2 \times 10^{-6})}{(0.5)^2} \]

\[ F = 8.99 \times 10^9 \times \frac{6 \times 10^{-12}}{0.25} = 8.99 \times 10^9 \times 2.4 \times 10^{-11} = 0.21576 \ \text{N} \]

The force is attractive because the charges are opposite.

Practice Set

• Level 1: What is the nature of force between two positive charges?
• Level 2: Two charges of +5 \( \mu \)C and +10 \( \mu \)C are placed 1 m apart. Calculate the force between them.
• Level 3: Three charges +2 \( \mu \)C, -3 \( \mu \)C, and +4 \( \mu \)C are placed at the vertices of an equilateral triangle of side 0.5 m. Calculate the net force on the +2 \( \mu \)C charge.

Answer Key

Level 1: The force is repulsive.

Level 2:

\[ F = k \frac{q_1 q_2}{r^2} = 8.99 \times 10^9 \times \frac{(5 \times 10^{-6})(10 \times 10^{-6})}{1^2} = 0.4495 \ \text{N} \]

Force is repulsive.

Level 3: Calculate forces due to each charge on +2 \( \mu \)C charge using Coulomb's law and vector addition. (Detailed vector calculations required.)

Electric Field and Dipole

Electric Field

The electric field \( \vec{E} \) at a point in space is defined as the force \( \vec{F} \) experienced by a positive test charge \( q_0 \) placed at that point divided by the magnitude of the test charge:

\[ \vec{E} = \frac{\vec{F}}{q_0} \]

It is a vector quantity, with direction of force on a positive charge.

Electric Field Due to a Point Charge

The magnitude of the electric field due to a point charge \( q \) at a distance \( r \) is:

\[ E = \frac{1}{4 \pi \varepsilon_0} \frac{|q|}{r^2} \]

The direction is radially outward for positive charge and inward for negative charge.

Electric Field Lines

Electric field lines are imaginary lines representing the direction of the electric field. They start from positive charges and end on negative charges, never intersect, and their density indicates field strength.

Electric Dipole

An electric dipole consists of two equal and opposite charges separated by a small distance \( 2a \). The dipole moment \( \vec{p} \) is defined as:

\[ \vec{p} = q \times 2a \ \hat{n} \]

where \( \hat{n} \) is the unit vector from negative to positive charge.

Electric Field Due to a Dipole

At a point on the axial line (along the dipole axis) at distance \( r \) (where \( r \gg a \)):

\[ E_{axial} = \frac{1}{4 \pi \varepsilon_0} \frac{2p}{r^3} \]

At a point on the equatorial line (perpendicular bisector) at distance \( r \):

\[ E_{equatorial} = \frac{1}{4 \pi \varepsilon_0} \frac{p}{r^3} \]

Torque on a Dipole in an Electric Field

A dipole in a uniform electric field experiences a torque \( \tau \) given by:

\[ \vec{\tau} = \vec{p} \times \vec{E} \]

The magnitude is \( \tau = p E \sin \theta \), where \( \theta \) is the angle between \( \vec{p} \) and \( \vec{E} \).

Solved Examples

Example 2: Calculate the electric field at a point 0.1 m away from a dipole with dipole moment \( 5 \times 10^{-8} \) C·m on the axial line.

Solution:

Given: \( p = 5 \times 10^{-8} \) C·m, \( r = 0.1 \) m

\[ E = \frac{1}{4 \pi \varepsilon_0} \frac{2p}{r^3} = 9 \times 10^9 \times \frac{2 \times 5 \times 10^{-8}}{(0.1)^3} = 9 \times 10^9 \times \frac{10 \times 10^{-8}}{10^{-3}} = 9 \times 10^9 \times 10^{-4} = 9 \times 10^5 \ \text{N/C} \]

Practice Set

• Level 1: What is the direction of electric field due to a positive point charge?
• Level 2: Calculate the electric field at a point 0.2 m away from a charge of +4 \( \mu \)C.
• Level 3: A dipole with dipole moment \( 3 \times 10^{-8} \) C·m is placed in a uniform electric field of 1000 N/C at an angle of 60°. Calculate the torque on the dipole.

Answer Key

Level 1: The electric field points radially outward from the positive charge.

Level 2:

\[ E = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2} = 9 \times 10^9 \times \frac{4 \times 10^{-6}}{(0.2)^2} = 9 \times 10^9 \times \frac{4 \times 10^{-6}}{0.04} = 9 \times 10^9 \times 10^{-4} = 9 \times 10^5 \ \text{N/C} \]

Level 3:

\[ \tau = p E \sin \theta = 3 \times 10^{-8} \times 1000 \times \sin 60^\circ = 3 \times 10^{-5} \times 0.866 = 2.598 \times 10^{-5} \ \text{N m} \]

Gauss's Theorem and Applications

Electric Flux

Electric flux \( \Phi_E \) through a surface is the measure of the number of electric field lines passing through that surface. For a uniform electric field \( E \) and surface area \( A \) inclined at angle \( \theta \),

\[ \Phi_E = E A \cos \theta \]

Gauss's Theorem

Gauss's theorem states that the net electric flux through any closed surface (Gaussian surface) is equal to the net charge enclosed \( q_{enc} \) divided by the permittivity of free space \( \varepsilon_0 \):

\[ \oint \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\varepsilon_0} \]

Applications of Gauss's Law

• Charged Spherical Shell: Electric field outside is \( E = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2} \), inside is zero.
• Infinite Charged Plane: Electric field is constant and given by \( E = \frac{\sigma}{2 \varepsilon_0} \), where \( \sigma \) is surface charge density.
• Long Charged Wire: Electric field at distance \( r \) is \( E = \frac{\lambda}{2 \pi \varepsilon_0 r} \), where \( \lambda \) is linear charge density.

Solved Examples

Example 3: Calculate the electric flux through a cube of side 0.1 m placed in an electric field \( E_x = 800 x^{1/2} \) N/C, where \( x \) is in meters.

Solution:

Calculate flux through faces perpendicular to x-axis:

At \( x = 0.1 \), \( E = 800 \times (0.1)^{1/2} = 800 \times 0.3162 = 253 \) N/C

Area \( A = (0.1)^2 = 0.01 \) m²

Flux through face at \( x=0.1 \): \( \Phi = E A = 253 \times 0.01 = 2.53 \) Nm²/C

Flux through face at \( x=0 \) is zero (since \( E_x = 0 \) at \( x=0 \)).

Total flux \( \Phi = 2.53 \) Nm²/C

Using Gauss's law, charge enclosed \( q = \varepsilon_0 \Phi = 8.854 \times 10^{-12} \times 2.53 = 2.24 \times 10^{-11} \) C

Practice Set

• Level 1: What is the electric flux through a surface perpendicular to a uniform electric field of 100 N/C and area 0.5 m²?
• Level 2: Calculate the electric field at a distance 0.2 m from a long charged wire with linear charge density \( 5 \times 10^{-6} \) C/m.
• Level 3: A spherical shell of radius 0.1 m carries a charge of 4 \( \mu \)C. Calculate the electric field at a point 0.05 m from the center inside the shell and at 0.15 m outside the shell.

Answer Key

Level 1: \( \Phi = E A = 100 \times 0.5 = 50 \) Nm²/C

Level 2:

\[ E = \frac{\lambda}{2 \pi \varepsilon_0 r} = \frac{5 \times 10^{-6}}{2 \pi \times 8.854 \times 10^{-12} \times 0.2} = 4.49 \times 10^6 \ \text{N/C} \]

Level 3:

Inside shell (r = 0.05 m): \( E = 0 \)

Outside shell (r = 0.15 m):

\[ E = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2} = 9 \times 10^9 \times \frac{4 \times 10^{-6}}{(0.15)^2} = 1.6 \times 10^6 \ \text{N/C} \]

Quick Reference Table

Coulomb's Law: \( F = \frac{1}{4 \pi \varepsilon_0} \frac{|q_1 q_2|}{r^2} \)

Electric Field of Point Charge: \( E = \frac{1}{4 \pi \varepsilon_0} \frac{|q|}{r^2} \)

Electric Dipole Moment: \( \vec{p} = q \times 2a \)

Electric Field of Dipole (Axial): \( E = \frac{1}{4 \pi \varepsilon_0} \frac{2p}{r^3} \)

Electric Field of Dipole (Equatorial): \( E = \frac{1}{4 \pi \varepsilon_0} \frac{p}{r^3} \)

Torque on Dipole: \( \vec{\tau} = \vec{p} \times \vec{E} \)

Electric Flux: \( \Phi_E = E A \cos \theta \)

Gauss's Law: \( \oint \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\varepsilon_0} \)

Electric Field of Charged Shell (Outside): \( E = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2} \)

Electric Field of Charged Shell (Inside): \( E = 0 \)

Electric Field of Infinite Wire: \( E = \frac{\lambda}{2 \pi \varepsilon_0 r} \)

Common Mistakes and Misconceptions

• Confusing electric field with electric potential; electric field is force per unit charge, potential is work done per unit charge.
• Ignoring vector nature of forces in Coulomb's law; forces must be added vectorially.
• Neglecting units; always use SI units (meters, coulombs, newtons).
• Misapplying Gauss's law without considering symmetry of the Gaussian surface.
• Assuming electric field inside a conductor is non-zero; it is zero in electrostatic equilibrium.
• Incorrect sign conventions; positive charges repel, negative charges attract.

Glossary

• Electric Charge: A property of matter causing it to experience force in an electric field.
• Coulomb's Law: Law describing force between two point charges.
• Electric Field: A vector field representing force per unit positive charge at a point.
• Electric Dipole: Two equal and opposite charges separated by a distance.
• Dipole Moment: A vector quantity representing strength and orientation of a dipole.
• Electric Flux: Measure of electric field lines passing through a surface.
• Gauss's Law: Relates electric flux through a closed surface to enclosed charge.
• Permittivity of Free Space (\( \varepsilon_0 \)): A constant describing electric properties of vacuum.
• Torque: A measure of the turning force on an object.