To find the electric field intensity along the $z$-axis at a distance $z$ from the origin, due to a uniformly charged disc of radius $r$ with surface charge density $\sigma$, we can break the problem into the following steps:

1. Symmetry and Setup:

The disc is uniformly charged and lies in the $xy$-plane with its center at the origin. Due to the symmetry of the disc, the electric field along the $z$-axis will point along the $z$-direction.

2. Small Element of Charge:

Consider a small ring of radius $\rho$ and thickness $d\rho$ on the disc. The area of this ring is $dA = 2\pi \rho \, d\rho$. The charge on this ring is:

$dq = \sigma \, dA = \sigma \, 2\pi \rho \, d\rho$

3. Electric Field Due to a Ring Element:

The electric field at a point on the $z$-axis due to this ring is composed of two components:

• One in the $z$-direction.
• One in the radial direction, which will cancel out due to symmetry when integrated over the whole disc.

The vertical component of the electric field is:

$dE_z = \frac{1}{4\pi \epsilon_0} \cdot \frac{dq \cdot z}{(z^2 + \rho^2)^{3/2}}$

Substitute $dq$:

$dE_z = \frac{1}{4\pi \epsilon_0} \cdot \frac{\sigma \, 2\pi \rho \, z \, d\rho}{(z^2 + \rho^2)^{3/2}}$

4. Total Electric Field:

To find the total electric field, integrate $dE_z$ from $\rho = 0$ to $\rho = r$:

$E_z = \int_0^r \frac{1}{4\pi \epsilon_0} \cdot \frac{\sigma \, 2\pi \rho \, z}{(z^2 + \rho^2)^{3/2}} \, d\rho$

Simplifying:

$E_z = \frac{\sigma z}{2\epsilon_0} \int_0^r \frac{\rho \, d\rho}{(z^2 + \rho^2)^{3/2}}$

5. Solving the Integral:

The integral can be solved using a substitution. Let $u = z^2 + \rho^2$, which gives $du = 2\rho d\rho$. The limits of integration change as follows:

• When $\rho = 0$, $u = z^2$.
• When $\rho = r$, $u = z^2 + r^2$.

Thus, the integral becomes:

$\int_0^r \frac{\rho \, d\rho}{(z^2 + \rho^2)^{3/2}} = \int_{z^2}^{z^2 + r^2} \frac{du}{2u^{3/2}}$

This evaluates to:

$\left[ -\frac{1}{\sqrt{u}} \right]_{z^2}^{z^2 + r^2} = \frac{1}{z} - \frac{1}{\sqrt{z^2 + r^2}}$

6. Final Expression for the Electric Field:

Substituting this result back into the expression for $E_z$, we get:

$E_z = \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{z}{\sqrt{z^2 + r^2}} \right)$

Conclusion:

The electric field intensity along the $z$-axis at a distance $z$ from the origin due to the uniformly charged disc is:

$E_z = \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{z}{\sqrt{z^2 + r^2}} \right)$

Would you like further explanation or clarification?

Here are some follow-up questions:

1. How does the electric field behave as $z$ approaches zero?
2. What happens to the electric field when $z \gg r$?
3. How would the field change if the disc were a ring instead of a solid disc?
4. How does the surface charge density $\sigma$ affect the strength of the field?
5. Can this result be extended to non-uniformly charged discs?

Tip: The electric field due to a uniformly charged disc along its axis diminishes as you move far from the disc, approaching the behavior of a point charge.