Knowlet

Unit 1: Electric Field and Electric Potential

1. Electric Field and Field Lines

An Electric Field (E) is a vector field surrounding an electric charge that exerts a force on other charges. It is defined as the force (F) experienced by a small positive test charge (q0) divided by the charge itself.

Formula: E = F / q0

The unit of the electric field is Newtons per Coulomb (N/C) or Volts per meter (V/m).

Electric Field Lines

Electric field lines are imaginary lines used to visualize the electric field.

  • They originate from positive charges and terminate on negative charges (or at infinity).
  • The tangent to a field line at any point gives the direction of the electric field E at that point.
  • The density of the field lines (how close they are) in a region is proportional to the strength (magnitude) of the electric field.
  • Electric field lines can never cross each other. If they did, it would imply two different directions of the E-field at the same point, which is impossible.

2. Electric Flux (ΦE)

Electric flux is the measure of the "flow" of the electric field through a given surface. It quantifies how many electric field lines pass through a specific area.

For a uniform electric field E passing through a flat area A with an angle θ between E and the area's normal vector:

Formula (Uniform Field): ΦE = E A cos(θ)

In general, for a non-uniform field and a curved surface, the flux is calculated by integrating the dot product of the electric field (E) and the differential area vector (dA) over the entire surface.

Formula (General): ΦE = ∫S E ⋅ dA

3. Gauss's Law and its Applications

Gauss's Law relates the net electric flux (ΦE) through a closed surface (called a "Gaussian surface") to the net electric charge (Qenclosed) enclosed within that surface.

Gauss's Law: ΦE = ∮ E ⋅ dA = Qenclosed / ε0

Where ε0 is the permittivity of free space (8.854 × 10-12 C2/N·m2).

Exam Tip: Gauss's Law is most useful for calculating the electric field in situations with high symmetry (spherical, cylindrical, or planar). The key is to choose a Gaussian surface that matches the symmetry, so that the electric field E is constant and perpendicular to the surface.

Applications of Gauss's Law

a) Spherical Symmetry (e.g., Uniformly Charged Sphere)

To find the E-field from a sphere of radius R and total charge Q, we use a spherical Gaussian surface of radius r.

  • Outside the sphere (r > R): The Gaussian surface encloses all the charge Q.

    E(4πr2) = Q / ε0 => E = Q / (4πε0r2)

    (The field is identical to that of a point charge Q at the center).

  • Inside a conducting sphere (r < R): Qenclosed = 0 (charge is on the surface). Therefore, E = 0.

b) Cylindrical Symmetry (e.g., Infinite Line of Charge)

For a line with charge per unit length λ, we use a cylindrical Gaussian surface of radius r and length L.

The flux is only through the curved wall (E is parallel to the end caps). Qenclosed = λL.

E(2πrL) = λL / ε0 => E = λ / (2πε0r)

c) Planar Symmetry (e.g., Infinite Sheet of Charge)

For a sheet with charge per unit area σ, we use a cylindrical "pillbox" Gaussian surface that pokes through the sheet.

Flux is through the two end caps (Area A). Qenclosed = σA.

E(A) + E(A) = σA / ε0 => 2EA = σA / ε0 => E = σ / (2ε0)

Note: The field is constant and does not depend on the distance from the sheet.

4. Conservative Nature of Electrostatic Field

An electrostatic field is a conservative field. This means the work done by the field in moving a charge from one point to another is independent of the path taken.

A direct consequence of this is that the work done in moving a charge around any closed loop is zero.

∮ E ⋅ dl = 0

Mathematically, this is equivalent to saying the curl of the electrostatic field is zero:

∇ × E = 0

5. Electrostatic Potential (V)

Because the electrostatic field is conservative, we can define a scalar quantity called Electrostatic Potential (V). It is defined as the potential energy (U) per unit charge (q).

Definition: V = U / q

The potential difference (ΔV) between two points is the work done per unit charge (W) to move a charge from one point to the other.

ΔV = Vb - Va = -∫ab E ⋅ dl

The unit of potential is the Volt (V), where 1 Volt = 1 Joule / Coulomb.

6. Relation between E-Field and Potential

The electric field E is the negative gradient of the potential V. The gradient is a vector operator (∇).

Formula: E = -∇V

This means the E-field points in the direction of the steepest decrease in potential.

In Cartesian coordinates, this breaks down into components:

  • Ex = -∂V/∂x
  • Ey = -∂V/∂y
  • Ez = -∂V/∂z

7. Laplace's and Poisson's Equations

These two equations are fundamental in electrostatics. They are derived by combining Gauss's Law (in differential form) with the E-V relation.

  1. Gauss's Law (differential): ∇ ⋅ E = ρ / ε0 (where ρ is the volume charge density)
  2. E-V Relation: E = -∇V

Substitute (2) into (1):

∇ ⋅ (-∇V) = ρ / ε0 => -∇2V = ρ / ε0

This gives Poisson's Equation, which relates the potential in a region to the charge density in that region.

Poisson's Equation:2V = -ρ / ε0

In a region of space where there is no charge (ρ = 0), Poisson's equation simplifies to Laplace's Equation.

Laplace's Equation:2V = 0

The operator ∇2 is called the Laplacian.

8. The Uniqueness Theorem

The syllabus requires the statement only.

Uniqueness Theorem Statement: If the potential (V) or its normal derivative (∂V/∂n) is specified on all boundaries (conductors, surfaces at infinity, etc.) of a region that obeys Poisson's equation, then the solution for the potential (V) within that region is unique.
What this means: It doesn't matter *how* you find a solution (e.g., "guessing", separation of variables, method of images). If your solution (1) satisfies Poisson's/Laplace's equation and (2) matches all the boundary conditions, it is the *only* correct solution.

9. Electric Dipole

An electric dipole consists of two equal and opposite charges (+q and -q) separated by a small vector distance (d, pointing from -q to +q).

The electric dipole moment (p) is a vector:

Formula: p = qd

Potential and Electric Field of a Dipole

At a point P far away from the dipole (r >> d), the potential V and field E can be calculated. In spherical coordinates (r, θ), where θ is the angle from the dipole axis:

  • Potential (V): V(r, θ) = (p cos(θ)) / (4πε0r2)
  • Electric Field (E): Derived from E = -∇V. It has two components:
    • Radial component (Er): Er = (2p cos(θ)) / (4πε0r3)
    • Tangential component (Eθ): Eθ = (p sin(θ)) / (4πε0r3)

Force and Torque on a Dipole

When a dipole (p) is placed in an external electric field (E):

  • In a UNIFORM E-field:
    • The net force on the dipole is zero.
    • It experiences a torque (τ) that tries to align it with the field.
    • Torque Formula: τ = p × E (Magnitude: τ = pE sin(θ))
    • Potential Energy (U): U = -p ⋅ E = -pE cos(θ)
  • In a NON-UNIFORM E-field:
    • The dipole experiences both a torque *and* a net force.
    • The force is given by F = (p ⋅ ∇)E.

Did this resource help you study?

Share feedback or report issues to help improve this resource.