Electric Charges and Fields
Unit I: Electrostatics (09 marks) – Physics for IIT JEE, NDA, and Other Competitive Exams
Introduction to Electric Charges
Electric charge is a fundamental property of matter responsible for electric phenomena. It is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. Electric charge exists in two forms, positive and negative, commonly carried by protons and electrons respectively.
Key Points
- Electric charge is quantized, meaning it exists in discrete units
- The smallest unit of charge is the elementary charge, e = 1.602 × 10-19 Coulomb
- Like charges repel, unlike charges attract
- Charge is always conserved in isolated systems
Types of Electric Charges
There are two types of electric charges:
| Property | Positive Charge | Negative Charge |
|---|---|---|
| Carried by | Protons | Electrons |
| Magnitude | +e | -e |
| Interaction | Like charges repel; unlike charges attract | |
Conservation of Charge
The principle of conservation of charge states that the net electric charge in an isolated system remains constant over time. Electric charge cannot be created or destroyed; it can only be transferred from one object to another.
Example: Conservation of Charge
Consider charging a rod by rubbing it with silk. Before rubbing, both the rod and silk are electrically neutral (net charge = 0). After rubbing, the rod gains a negative charge of -q, while the silk gains a positive charge of +q. The total charge of the system before and after remains zero, demonstrating charge conservation.
Coulomb’s Law
Coulomb’s law quantifies the electrostatic force between two point charges. It states that the magnitude of the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.
Where:
- F is the electrostatic force between charges (in Newtons)
- q₁ and q₂ are the magnitudes of the charges (in Coulombs)
- r is the distance between the charges (in meters)
- k is Coulomb’s constant (9 × 10⁹ N·m²/C²)
- ε₀ is the permittivity of free space (8.85 × 10⁻¹² C²/N·m²)
The direction of the force is along the line joining the two charges. If the charges are of the same sign (both positive or both negative), the force is repulsive. If the charges are of opposite signs, the force is attractive.
Practice Problem 1
Two point charges q₁ = 3 μC and q₂ = -4 μC are placed 30 cm apart in vacuum. Calculate the magnitude and direction of the electrostatic force between them.
Solution:
Using Coulomb’s law: F = k·|q₁q₂|/r²
F = (9 × 10⁹ N·m²/C²) × |(3 × 10⁻⁶ C) × (-4 × 10⁻⁶ C)|/(0.3 m)²
F = (9 × 10⁹) × (12 × 10⁻¹²)/(0.09)
F = (9 × 12 × 10⁻³)/(0.09)
F = 108 × 10⁻³/0.09 = 1.2 N
Since the charges are of opposite signs (positive and negative), the force is attractive, directed along the line joining the charges.
Forces Between Multiple Charges: The Superposition Principle
When multiple charges are present, the net force on any charge is the vector sum of all individual forces exerted by other charges. This is known as the superposition principle.
This principle is particularly important for solving problems involving more than two charges, where the total force must be calculated using vector addition.
Example: Superposition Principle
Consider three charges placed on the x-axis: q₁ = 2 μC at x = 0, q₂ = -3 μC at x = 3 m, and q₃ = 4 μC at x = 6 m. To find the net force on q₃, we:
- Calculate the force F₁₃ due to q₁ on q₃ (attractive)
- Calculate the force F₂₃ due to q₂ on q₃ (repulsive)
- Find the vector sum: Fnet = F₁₃ + F₂₃
Continuous Charge Distribution
In many practical situations, charge is distributed continuously over a body rather than as discrete point charges. For such cases, we divide the charge distribution into infinitesimal elements and integrate to find the total effect.
Depending on the geometry, charge distributions can be:
- Linear (charge per unit length, λ)
- Surface (charge per unit area, σ)
- Volume (charge per unit volume, ρ)
For a linear charge distribution: dq = λ dl
For a surface charge distribution: dq = σ dA
For a volume charge distribution: dq = ρ dV
Electric Field
The electric field is a vector quantity that represents the force per unit charge experienced by a test charge at any point in space due to the presence of other charges.
Where E is the electric field, F is the force on a test charge q₀. The SI unit of electric field is Newton per Coulomb (N/C) or Volt per meter (V/m).
Electric Field Due to a Point Charge
The electric field at a distance r from a point charge q is given by:
The direction of the electric field is:
- Radially outward from a positive charge
- Radially inward toward a negative charge
Practice Problem 2
A charge of 5 μC is placed at the origin. Calculate the electric field at a point (0, 3) m.
Solution:
The distance from the origin to the point (0, 3) is r = 3 m.
Using the formula: E = k · q/r²
E = (9 × 10⁹ N·m²/C²) × (5 × 10⁻⁶ C)/(3 m)²
E = (9 × 10⁹) × (5 × 10⁻⁶)/9
E = 5 × 10³ N/C
Since the charge is positive, the electric field is directed radially outward from the origin, so it points along the positive y-axis.
Electric Field Lines
Electric field lines are imaginary lines that help visualize the electric field in a region. They represent the path that a positive test charge would follow if free to move.
Properties of electric field lines:
- They start from positive charges and end on negative charges
- They never cross each other
- The density of lines is proportional to the strength of the electric field
- They are perpendicular to equipotential surfaces
- For an isolated charge, they are radial
Electric Dipole
An electric dipole consists of two equal and opposite charges separated by a small distance. It is characterized by its dipole moment.
Where p is the dipole moment vector, q is the magnitude of each charge, and 2a is the distance between the charges. The direction of p is from the negative charge to the positive charge. The SI unit of dipole moment is Coulomb-meter (C·m).
Electric Field Due to a Dipole
The electric field due to a dipole varies with position. At a point along the axis of the dipole (axial position), at a distance r from the center of the dipole (where r >> a):
At a point on the perpendicular bisector of the dipole (equatorial position), at a distance r from the center (where r >> a):
Example: Electric Field of a Dipole
An electric dipole consists of charges +2 μC and -2 μC separated by 2 cm. Calculate:
- The dipole moment
- The electric field at a point 10 cm away along the axis of the dipole
a) Dipole moment:
p = q × 2a = 2 × 10⁻⁶ C × 0.02 m = 4 × 10⁻⁸ C·m
b) Electric field along the axis:
E = (1/4πε₀) × (2p/r³)
E = (9 × 10⁹) × (2 × 4 × 10⁻⁸)/(0.1)³
E = 9 × 10⁹ × 8 × 10⁻⁸ × 1000
E = 7.2 × 10⁵ N/C
Torque on a Dipole in a Uniform Electric Field
When an electric dipole is placed in a uniform electric field, it experiences a torque that tends to align the dipole moment with the field direction.
Where τ is the torque, p is the dipole moment, E is the electric field, and θ is the angle between p and E. The dipole is in stable equilibrium when aligned with the field (θ = 0°) and in unstable equilibrium when anti-aligned (θ = 180°).
Practice Problem 3
An electric dipole with a dipole moment of 3 × 10⁻²⁹ C·m is placed in a uniform electric field of 5 × 10⁴ N/C. If the dipole makes an angle of 30° with the field, calculate:
- The torque on the dipole
- The potential energy of the dipole
Solution:
a) Torque:
τ = p·E·sin θ
τ = (3 × 10⁻²⁹ C·m) × (5 × 10⁴ N/C) × sin(30°)
τ = 15 × 10⁻²⁵ × 0.5
τ = 7.5 × 10⁻²⁵ N·m
b) Potential energy:
U = -p·E·cos θ
U = -(3 × 10⁻²⁹) × (5 × 10⁴) × cos(30°)
U = -15 × 10⁻²⁵ × 0.866
U = -13 × 10⁻²⁵ J
Electric Flux
Electric flux is a measure of the electric field passing through a given area. It is a scalar quantity defined as the dot product of the electric field and the area vector.
Where Φₑ is the electric flux, E is the electric field, A is the area, and θ is the angle between the electric field and the normal to the area. The SI unit of electric flux is N·m²/C or V·m.
For a non-uniform electric field or curved surface, the flux is calculated by integration:
Gauss’s Theorem
Gauss’s law states that the total electric flux through a closed surface is equal to the total charge enclosed by the surface divided by the permittivity of free space.
This is one of Maxwell’s equations and is particularly useful for calculating electric fields in systems with symmetry.
Applications of Gauss’s Law
1. Field Due to an Infinitely Long Straight Wire
For a wire with uniform linear charge density λ, the electric field at a distance r from the wire is:
The field is radially outward from the wire if λ is positive and inward if λ is negative.
2. Field Due to a Uniformly Charged Infinite Plane Sheet
For a plane with uniform surface charge density σ, the electric field at any point is:
The field is perpendicular to the plane, directed away from the plane if σ is positive and toward the plane if σ is negative. Notably, the field is independent of the distance from the plane.
3. Field Due to a Uniformly Charged Thin Spherical Shell
For a spherical shell of radius R with uniform surface charge density σ, the electric field is:
Inside the shell (r < R): E = 0
Outside the shell (r > R): E = (1/4πε₀) × Q/r² = kQ/r²
Where Q is the total charge on the shell, Q = 4πR²σ. This shows that for points outside the shell, the field is the same as if all the charge were concentrated at the center.
Practice Problem 4
A thin spherical shell of radius 10 cm has a uniform surface charge density of 5 × 10⁻⁶ C/m². Calculate:
- The total charge on the shell
- The electric field at a point 5 cm from the center of the shell
- The electric field at a point 15 cm from the center of the shell
Solution:
a) Total charge:
Q = 4πR²σ
Q = 4π × (0.1 m)² × (5 × 10⁻⁶ C/m²)
Q = 4π × 10⁻² × 5 × 10⁻⁶
Q = 6.28 × 10⁻⁷ C
b) Electric field at r = 5 cm:
Since r < R (inside the shell), E = 0
c) Electric field at r = 15 cm:
Since r > R (outside the shell),
E = kQ/r²
E = (9 × 10⁹) × (6.28 × 10⁻⁷)/(0.15)²
E = 9 × 10⁹ × 6.28 × 10⁻⁷ / 0.0225
E = 2.51 × 10⁵ N/C
Important Formulas to Remember
- Coulomb’s Law: F = k·|q₁q₂|/r²
- Electric Field Due to Point Charge: E = kq/r²
- Electric Dipole Moment: p = q·2a
- Torque on Dipole: τ = p×E = p·E·sin θ
- Electric Flux: Φₑ = E·A = E·A·cos θ
- Gauss’s Law: ∮E·dA = qenclosed/ε₀
- Field Due to Infinite Wire: E = λ/(2πε₀r)
- Field Due to Infinite Plane: E = σ/(2ε₀)
- Field Due to Spherical Shell (outside): E = kQ/r²
- Field Due to Spherical Shell (inside): E = 0
Exam Tips for Electric Charges and Fields
For IIT JEE
- Focus on vector nature of electric fields and forces
- Practice problems involving multiple charges in different configurations
- Master Gauss’s law applications for symmetric charge distributions
- Understand the mathematical derivations of key formulas
For NDA
- Focus on the conceptual understanding of electric field and flux
- Remember the key formulas and their applications
- Practice numerical problems with different units
- Understand the physical interpretation of electric field lines
Common Mistakes to Avoid
- Forgetting to convert units (e.g., micro-coulombs to coulombs)
- Applying Gauss’s law without proper symmetry considerations
- Incorrect vector addition when calculating net forces
- Mixing up the formulas for electric field at axial and equatorial points of a dipole
- Neglecting the sign of charges when determining the direction of forces or fields
