Uniformly charged disc of radius r having self is charged density Sigma is placed in xy plane with its centre at origin find the electric field intensity along z Axis at a distance z from the origin

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.

Uniformly charged disc of radius r having surface charge density Sigma is placed in xy plane with its centre at origin. Find the electric field intensity along z-axis at a distance z from the origin.

Learn how to calculate the electric field intensity along the z-axis at a distance z from the origin due to a uniformly charged disc with radius r and surface charge density σ. This detailed solution involves integration and applies concepts from electrostatics. Suitable for undergraduate physics students.

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