Unit 1: Electrostatics

1. Electrostatic Field & Electric Flux

Electrostatic Field

An electrostatic field is a region of space around a charged particle or source charge where another charged particle experiences an electrostatic force. The intensity of this field at any point is defined as the force experienced per unit positive test charge placed at that point.

Electric Field Intensity (E):
E = F / q₀

Where F is the electrostatic force and q₀ is an infinitesimally small positive test charge. The SI unit of electric field intensity is Newtons per Coulomb (N/C) or Volts per meter (V/m).

Electric Flux

Electric flux (represented by Φ) is a measure of the total number of electric field lines passing through a given surface area. Mathematically, the electric flux through an infinitesimal area element dA is given by the dot product of the electric field vector and the area vector.

dΦ = E • dA = E dA cos(θ)

Where θ is the angle between the electric field vector E and the area vector dA (which points perpendicular to the surface). For a closed surface, the total electric flux is the surface integral over the entire area:

Φ = ∮ E • dA

2. Gauss's Theorem of Electrostatics

Gauss's Theorem is a fundamental law in electromagnetism that relates the spatial distribution of electric charge to the resulting electric field.

Statement of Gauss's Theorem:
The net electric flux through any closed hypothetical surface (referred to as a Gaussian surface) is equal to 1/ε₀ times the net charge enclosed within that surface.

Mathematical Equation:
Φ = ∮ E • dA = q_enclosed / ε₀

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

Important Observations

• The flux depends only on the total charge enclosed within the surface, not on the distribution of charges.
• Charges outside the closed surface contribute zero net flux through the surface because any field line entering the surface must also exit it.
• The choice of the Gaussian surface is arbitrary, but using surfaces that match the symmetry of the charge distribution (spherical, cylindrical, or planar) makes solving the integrals straightforward.

3. Applications of Gauss's Theorem

Application 1: Electric Field due to a Point Charge

To find the electric field at a distance r from a point charge q:

1. Construct a concentric spherical Gaussian surface of radius r centered on the charge.
2. By spherical symmetry, the electric field intensity E is radial and has the same magnitude at every point on the surface. The angle θ between E and dA is 0° at all points.
3. Apply Gauss's Theorem:

   ∮ E • dA = ∮ E dA cos(0°) = E ∮ dA = E (4πr²) = q / ε₀

4. Solve for E:

   E = q / (4πε₀ r²)

This result is identical to the expression derived from Coulomb's Law, validating Gauss's Theorem.

Application 2: Electric Field due to an Infinite Line of Charge

Consider an infinitely long, straight line with a uniform linear charge density λ (charge per unit length).

1. Construct a coaxial cylindrical Gaussian surface of radius r and length L.
2. The total surface consists of three parts: two flat circular end caps and one curved cylindrical surface.
3. The electric field points radially outwards. Therefore, E is parallel to the area vector of the curved surface (θ = 0°) and perpendicular to the area vectors of the flat end caps (θ = 90°). Thus, flux through the end caps is zero.
4. Apply Gauss's Theorem:

   ∮ E • dA = E (2πrL) = q_enclosed / ε₀ = λL / ε₀

5. Solve for E:

   E = λ / (2πε₀ r)

Application 3: Electric Field due to a Uniformly Charged Spherical Shell

Consider a thin spherical shell of radius R carrying a total charge Q uniformly distributed over its surface. The surface charge density is σ = Q / (4πR²).

• Case 1: At an external point (r > R)
  Construct a spherical Gaussian surface of radius r concentric with the shell.

  E (4πr²) = Q / ε₀ → E = Q / (4πε₀ r²)

  Observation: For points outside the shell, the entire charge behaves as if it is concentrated at the center.
• Case 2: On the surface (r = R)
  Replacing r with R in the external field formula:

  E = Q / (4πε₀ R²) = σ / ε₀

• Case 3: At an internal point (r < R)
  Construct a spherical Gaussian surface of radius r inside the shell. Since all charge resides on the outer surface of the shell, the enclosed charge is zero.

  E (4πr²) = 0 / ε₀ → E = 0

  Observation: The electrostatic field inside a uniformly charged conducting spherical shell is always zero. This is the underlying principle behind electrostatic shielding.

Application 4: Electric Field due to a Uniformly Charged Solid Sphere

Consider an insulating solid sphere of radius R with a total charge Q distributed uniformly throughout its volume. The volume charge density is ρ = Q / ((4/3)πR³).

• Case 1: At an external point (r ≥ R)
  Using a concentric spherical Gaussian surface of radius r:

  E (4πr²) = Q / ε₀ → E = Q / (4πε₀ r²)

• Case 2: At an internal point (r < R)
  Construct a spherical Gaussian surface of radius r inside the sphere. The charge enclosed within this smaller sphere is:

  q_enclosed = ρ × (4/3)πr³ = Q (r³ / R³)

  Applying Gauss's Theorem:

  E (4πr²) = [Q (r³ / R³)] / ε₀

  Solving for E yields:

  E = Q r / (4πε₀ R³) = ρ r / (3ε₀)

Application 5: Electric Field due to an Infinite Plane Charged Sheet

Consider an infinite thin plane sheet of charge with a uniform surface charge density σ.

1. By symmetry, the electric field must point normally away from the sheet on both sides.
2. Construct a cylindrical Gaussian "pillbox" of cross-sectional area A and total length 2r cutting perpendicularly through the sheet.
3. The field lines are parallel to the curved surface of the cylinder, so the flux through it is zero. The field lines are perpendicular to the two flat circular ends.
4. Apply Gauss's Theorem:

   ∮ E • dA = E A (left end) + E A (right end) = 2 E A = q_enclosed / ε₀

   Since the enclosed charge is σ A:

   2 E A = σ A / ε₀

5. Solve for E:

   E = σ / (2ε₀)

Observation: The electric field due to an infinite plane sheet of charge is completely independent of the distance r from the sheet.

Comparison of Electric Fields

4. Electric Potential as a Line Integral

The electrostatic field is conservative. This means the work done in moving a charge between two points in an electrostatic field is independent of the path taken.

Definition of Electric Potential (V):
The electric potential at any point in an electric field is the negative line integral of the electric field from infinity to that point.

V = - ∫_∞^r E • dl

Where dl is an infinitesimal displacement vector along the path of integration. The potential difference between two points A and B is given by:

V_B - V_A = - ∫_A^B E • dl

Common Mistake

Students often forget the negative sign in the line integral definition. The negative sign is critical because work must be done against the electrostatic force when bringing a positive test charge closer to a positive source charge.

5. Potential due to a Point Charge & Dipole

Potential due to a Point Charge

To find the electric potential V at a distance r from a positive point charge q:

1. Substitute the electric field of a point charge, E = q / (4πε₀ x²) r̂, into the line integral definition:

   V = - ∫_∞^r [ q / (4πε₀ x²) ] dx

2. Evaluate the integral:

   V = - (q / 4πε₀) [ -1/x ]_∞^r = q / (4πε₀ r)

Potential due to an Electric Dipole

An electric dipole consists of two equal and opposite charges, -q and +q, separated by a distance 2a. The electric dipole moment vector is defined as p = q × (2a), directed from -q to +q.

Let us calculate the electric potential at a point P situated at a distance r from the center of the dipole O, where the position vector r makes an angle θ with the dipole axis:

1. Let the distances of P from charges +q and -q be r₁ and r₂ respectively.
2. By the superposition principle:

   V = V₁ + V₂ = (1 / 4πε₀) × [ q / r₁ - q / r₂ ]

3. If r >> a, using geometry and approximations:

   r₁ ≈ r - a cos(θ)
   r₂ ≈ r + a cos(θ)

4. Substitute these values back into the potential equation:

   V ≈ (q / 4πε₀) × [ 1 / (r - a cos(θ)) - 1 / (r + a cos(θ)) ]
   
   V ≈ (q / 4πε₀) × [ (2a cos(θ)) / (r² - a² cos²(θ)) ]

5. Since r >> a, we can neglect a² cos²(θ) in the denominator. Substituting p = 2aq:

   V = p cos(θ) / (4πε₀ r²) = p • r̂ / (4πε₀ r²)

Special Cases for a Dipole

• On the axial line (θ = 0° or 180°):

  V = ± p / (4πε₀ r²)

• On the equatorial line (θ = 90°):

  V = p cos(90°) / (4πε₀ r²) = 0

6. Capacitance of an Isolated Spherical Conductor

Capacitance (C) is the capacity of a conductor to store electric charge and electrical energy. It is defined as the ratio of the charge Q supplied to the conductor to the increase in its electric potential V:

C = Q / V

Consider an isolated solid or hollow conducting sphere of radius R placed in a vacuum, carrying a charge Q. Since it is a conductor, the charge distributes itself uniformly on the outer surface. The electric potential on its surface is:

V = Q / (4πε₀ R)

Substituting this potential into the capacitance definition:

C = Q / [ Q / (4πε₀ R) ] = 4πε₀ R

Observation: The capacitance of an isolated spherical conductor is directly proportional to its radius (C ∝ R).

7. Condensers: Parallel Plate, Spherical, & Cylindrical

A condenser (or capacitor) consists of two conductors carrying equal and opposite charges, separated by an insulating medium (dielectric). Its purpose is to increase charge storage capacity by lowering the electric potential of the system.

1. Parallel Plate Condenser

Consists of two parallel conducting plates, each of area A, separated by a small distance d. Let the plates carry charges +Q and -Q respectively, yielding a surface charge density of magnitude σ = Q / A.

1. The electric field between the plates (ignoring edge effects) is uniform:

   E = σ / ε₀ = Q / (A ε₀)

2. The potential difference V between the plates is:

   V = E × d = Q d / (A ε₀)

3. Using C = Q / V:

   C = ε₀ A / d

2. Spherical Condenser

Consists of two concentric spherical conducting shells of inner radius a and outer radius b (where b > a).

• Case: Inner sphere is given a positive charge +Q, and the outer sphere is grounded (potential at b is zero).
• The potential difference V between the spheres is:

  V = V_a - V_b = [ Q / (4πε₀ a) ] - [ Q / (4πε₀ b) ]
  
  V = (Q / 4πε₀) × [ (b - a) / (ab) ]

• Thus, the capacitance is:

  C = Q / V = 4πε₀ ab / (b - a)

3. Cylindrical Condenser

Consists of two coaxial cylindrical conductors of length L and radii a (inner) and b (outer), where b > a. Let the inner cylinder carry a charge +Q (linear density λ = Q / L) and the outer cylinder be grounded.

1. The electric field at a radial distance r between the cylinders (a < r < b) is:

   E = λ / (2πε₀ r)

2. Calculate the potential difference V between the cylinders:

   V = - ∫_b^a E dr = ∫_a^b [ λ / (2πε₀ r) ] dr = (λ / 2πε₀) ln(b/a)

3. Substitute λ = Q / L into the expression for V:

   V = Q ln(b/a) / (2πε₀ L)

4. Thus, the capacitance is:

   C = Q / V = 2πε₀ L / ln(b/a)

Summary of Condensers

8. Energy Density in an Electrostatic Field

The process of charging a capacitor involves transferring charge from one plate to another against an opposing potential difference. This work is stored in the system as electrostatic potential energy.

Derivation of Stored Energy

1. At an intermediate stage of charging, let the charge on the capacitor be q and the potential difference be v = q / C.
2. The small work done dW in transferring an additional charge dq is:

   dW = v dq = (q / C) dq

3. The total work done W to charge the capacitor from 0 to Q is:

   U = W = ∫₀^Q (q / C) dq = Q² / (2C)

4. Using the relations Q = CV and C = Q/V, the total stored energy can be written as:

   U = (1/2) C V² = (1/2) Q V = Q² / (2C)

Energy Density (Energy per Unit Volume)

For a parallel plate capacitor, the volume enclosed between the plates is Volume = A × d. Also, we know that C = ε₀ A / d and the electric field between the plates is E = V / d (which means V = E d).

1. Substitute C and V into the energy equation:

   U = (1/2) × (ε₀ A / d) × (E d)² = (1/2) ε₀ E² (A d)

2. The electrostatic energy per unit volume (energy density, represented by u) is:

   u = U / (Volume) = U / (A d) = (1/2) ε₀ E²

Observation: Although derived using a parallel plate capacitor, the formula u = (1/2) ε₀ E² is a general result that holds true for any electrostatic field configuration in a vacuum.