Continuous Charge Distributions and Applications in Electrostatics

Surface Charge Density ( $\sigma$ ):
- Definition: Surface charge density is defined as the charge per unit area of a surface.
- Mathematical Expression: $"\sigma = \frac{dq}{ds}"$
- Variables: $dq$ represents the charge on an infinitesimal surface area $ds$ .
- Units: The SI units for $\sigma$ are coulomb/meter $^2$ ( $C/m^2$ or $C\,m^{-2}$ ).
- Examples:
  - Plane sheet of charge.
  - Conducting sphere.
Volume Charge Density ( $\rho$ ):
- Definition: Volume charge density is defined as the charge per unit volume.
- Mathematical Expression: $"\rho = \frac{dq}{dv}"$
- Variables: $dq$ represents the charge on an infinitesimal volume element $dv$ .
- Units: The SI units for $\rho$ are coulomb/meter $^3$ ( $C/m^3$ or $C\,m^{-3}$ ).
- Examples:
  - Charge on a dielectric sphere.
Linear Charge Distribution Examples (Implied):
- Charged straight wire.
- Circular charged ring.

Charge Residence: Charge given to a conductor always resides on its outer surface.
Uniform Distribution: If the surface of the conductor is uniform, the charge distributes itself uniformly across that surface.
Non-spherical Surfaces and Curvature:
- In conductors with non-spherical surfaces, the surface charge density ( $\sigma$ ) is not uniform.
- The surface charge density ( $\sigma$ ) is larger where the radius of curvature is small (i.e., at sharper points).
The Lightning Conductor:
- The functionality of a lightning conductor is based on the leakage of charge through sharp points.
- This leakage occurs due to the high surface charge density associated with the small radius of curvature at these sharp points.

Problem Statement: A ring of radius $R$ has a uniformly distributed charge $Q$ on it. A charge $q$ is subsequently placed at the centre of the ring. Find the increment in tension ( $\Delta T$ ) in the ring.
Solution Methodology:
- Consider an infinitesimal element of the ring subtending an angle $d\theta$ at the centre.
- The length of this element is $R\,d\theta$ .
- The elemental charge $dQ$ on this segment is calculated based on the total charge $Q$ and the circumference $2\pi R$ : $"dQ = \frac{Q}{2\pi R} (R\,d\theta) = \frac{Q\,d\theta}{2\pi}"$
- Force Balance (Equilibrium): For the equilibrium of this segment, we consider the outward repulsive force ( $F$ ) and the inward components of the tension increment ( $\Delta T$ ). $"F = 2\Delta T \sin\left(\frac{d\theta}{2}\right)"$
- Since the angle $d\theta$ is infinitesimal, we use the small angle approximation $"\sin\left(\frac{d\theta}{2}\right) \approx \frac{d\theta}{2}"$ : $"F = 2\Delta T \left(\frac{d\theta}{2}\right) = \Delta T\,d\theta"$
- Electric Repulsion Force ( $F$ ): The force between the central charge $q$ and the elemental charge $dQ$ is: $"F = \frac{1}{4\pi\epsilon_0} \frac{q\,dQ}{R^2}"$
- Combining Equations: Substituting the expression for $dQ$ into the force equation: $"\Delta T\,d\theta = \frac{1}{4\pi\epsilon_0} \frac{q}{R^2} \left(\frac{Q\,d\theta}{2\pi}\right)"$
- Final Result for Tension Increment: $"\Delta T = \frac{Qq}{8\pi^2\epsilon_0 R^2}"$

Problem Statement: A thin fixed ring of radius $a$ has a positive charge $q$ uniformly distributed over it. A particle of mass $m$ having a negative charge $-Q$ is placed on the axis of the ring at a distance $x$ from the centre, where $x \ll a$ . Show that the motion of the negatively charged particle is approximately simple harmonic (SHM) and calculate the time period of oscillation.
Preliminary Analysis:
- Consider an element $dq$ of the ring. The force on the point charge $-Q$ due to this element is: $"dF = \frac{1}{4\pi\epsilon_0} \frac{dq\,Q}{r^2}"$
- This force acts along the line $AB$ connecting the element and the point charge.
- Due to the symmetry of the ring, for every element $dq$ , there exists a diametrically situated element on the opposite side of the ring.
- The components of the force perpendicular to the axis will cancel out, leaving only the component directed toward the centre of the ring along the axis.