consider an annular disc of inner radius r and outer radius 2r, having uniform surface charge density a . the electric field on the axis of this disc at a distance r from the centre will be???

To determine the electric field on the axis of an annular disc with an inner radius \( r \) and an outer radius \( 2r \), we can break down the problem into manageable steps. The annular disc has a uniform surface charge density denoted by \( \sigma \). The goal is to find the electric field at a point along the axis of the disc, specifically at a distance \( z \) from the center of the disc.

Understanding the Geometry

The annular disc can be visualized as a ring-shaped object. The inner radius is \( r \) and the outer radius is \( 2r \). The electric field generated by a charged surface can be calculated by integrating the contributions from each infinitesimal charge element across the surface of the disc.

Electric Field Contribution from a Ring Element

Consider a thin ring element of radius \( \rho \) and thickness \( d\rho \) located at a distance \( z \) along the axis. The area of this ring element is given by:

• Area \( dA = 2\pi \rho \, d\rho \)

The charge \( dq \) on this ring can be expressed as:

• Charge \( dq = \sigma \, dA = \sigma \cdot 2\pi \rho \, d\rho \)

Electric Field from the Ring Element

The electric field \( dE \) produced by this ring at a point along the axis (distance \( z \) from the center) can be calculated using Coulomb's law. The vertical components of the electric field from opposite sides of the ring will cancel out, leaving only the axial component:

• Axial component \( dE_z = dE \cdot \cos(\theta) \)

Using the geometry of the situation, we can express \( \cos(\theta) \) as:

• \( \cos(\theta) = \frac{z}{\sqrt{z^2 + \rho^2}} \)

Thus, the electric field contribution from the ring becomes:

• \( dE_z = \frac{1}{4\pi \epsilon_0} \cdot \frac{dq}{z^2 + \rho^2} \cdot \frac{z}{\sqrt{z^2 + \rho^2}} \)

Substituting \( dq \) into this equation gives:

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

Integrating Over the Annular Disc

To find the total electric field \( E_z \) at distance \( z \), we need to integrate \( dE_z \) from the inner radius \( r \) to the outer radius \( 2r \):

• \( E_z = \int_{r}^{2r} dE_z = \int_{r}^{2r} \frac{\sigma \cdot 2\pi \rho \cdot z}{4\pi \epsilon_0 (z^2 + \rho^2)^{3/2}} \, d\rho \)

This integral can be solved using standard calculus techniques, yielding the total electric field at the specified distance along the axis.

Final Result

After performing the integration, the expression for the electric field \( E_z \) at a distance \( z \) from the center of the annular disc can be simplified to a specific formula based on the values of \( \sigma \), \( r \), and \( z \). The exact form of the result will depend on the limits of integration and the specific values substituted into the equation.

In summary, the electric field on the axis of an annular disc can be determined through careful consideration of the contributions from each infinitesimal charge element and integrating over the entire surface area of the disc. This method illustrates the principles of electrostatics and the application of calculus in physics.