Example: A disc of radius R has a uniform charge per unit area ċ. Calculate the electric field at a point P along the central axis of the disc at a distance x0 from its center.
You need to perform a Coulomb's-Law integration of the contribution of each surface area element of charge to the axial field. As a result of symmetry, the field at xo is directed along the central axis of symmetry. The integration over area can be performed as a one-dimensional integration over rings of radius r and width dr, with r running from 0 to R.
To calculate the electric field at point P along the central axis of the disc, we can use Gauss's Law. Gauss's Law relates the electric flux passing through a closed surface to the total charge enclosed by that surface.
Let's start by considering a Gaussian surface in the shape of a cylinder, with its axis coinciding with the central axis of the disc. The radius of this cylinder will be larger than the radius of the disc, R.
Since the disc has a uniform charge per unit area, we can consider an infinitesimally small, annular ring on its surface. The charge enclosed within this ring is given by:
dq = ċ * dA
where dq is the charge, ċ is the charge density (charge per unit area), and dA is the area of the annular ring.
The area of the annular ring, dA, is given by:
dA = 2π(r * dr)
where r is the radial distance from the center of the disc to the annular ring, and dr is an infinitesimally small thickness.
Now, let's integrate the charge over the entire disc. The total charge enclosed within the Gaussian surface is given by:
Q = ∫dq = ∫ƒã * dA
Integrating over the entire disc with radius R, we have:
Q = ∫₀ᴿ ∫₀²π(ƒã * 2π(r * dr)) dr
Simplifying this expression, we get:
Q = 2π²ƒã ∫₀ᴿ r * dr
Evaluating the integral, we find:
Q = π²ƒãR²
Applying Gauss's Law, the electric flux passing through the Gaussian surface is given by:
Φ = Q / ε₀
where ε₀ is the electric constant.
Since the electric field is constant along the cylindrical surface of the Gaussian surface, the electric flux through its curved surface is given by:
Φ = E * A
where E is the electric field and A is the area of the curved surface of the Gaussian surface.
The area of the curved surface of the Gaussian surface is given by:
A = 2πRL
where L is the length of the Gaussian surface.
Now, we can equate the expressions for the electric flux:
E * A = Q / ε₀
Substituting the expressions for A and Q, we get:
E * (2πRL) = π²ƒãR² / ε₀
Simplifying, we find:
E = ƒãR / (2ε₀)
Finally, we have the electric field at point P along the central axis of the disc at a distance x₀ from its center:
E = ƒãR / (2ε₀)