Flux passing through a disc with a concentric hole

Given a disc of radius R with a concentric hole of radius (3R)/4. The remaining portion of the disc is q/(n*epsilon_{0}). A charge q is placed at a distance R from the centre of the disc along its axis. We need to find the amount of flux passing through the disc.

Flux passing through a closed surface

The flux passing through a closed surface is given by the formula:

Φ = ∫∫ E . dA

Where E is the electric field, dA is the area vector, and the integral is taken over the entire surface.

Calculating electric field

To calculate the electric field, we can use Gauss's law:

Φ = ∫∫ E . dA = q/ε0

Where q is the charge enclosed by the surface and ε0 is the permittivity of free space.

Applying Gauss's law

We can apply Gauss's law by considering a cylindrical surface of radius r and length L, with its axis along the axis of the disc and passing through the charge q.

The electric field E is radial and has the same magnitude at every point on the cylindrical surface. The area vector dA is perpendicular to the surface and has a magnitude of 2πrL.

Hence, the flux passing through the cylindrical surface is given by:

Φ = E ∫∫ dA = E (2πrL)

By Gauss's law, Φ = q/ε0, so we can write:

E (2πrL) = q/ε0

Solving for E, we get:

E = q/(2πε0rL)

Finding flux through the disc

To find the flux passing through the disc, we need to integrate the electric field over the surface of the disc. Since the disc has a hole in the center, we need to integrate over two separate regions: the region inside the hole and the region outside the hole.

For the region outside the hole, the electric field is given by:

E = q/(2πε0rL)

Where r is the distance from the center of the disc to the point on the surface.

For the region inside the hole, the electric field is zero.

Hence, the flux passing through the disc is given by:

Φ = ∫∫ E . dA = ∫∫ Eoutside . dA + ∫∫ Einside . dA

Φ = ∫∫ Eoutside . dA

Φ = ∫∫ E . dA = ∫∫ q/(2πε0rL) dA

We can simplify this expression by using the fact that the disc is symmetric about its axis, so the electric field is also symmetric about its axis.

Hence, we can write:

Φ = q/(2πε0L) ∫∫ 1/r dA

The integral over the area is just the area of the disc, which is πR^2 - π(3R/4)^2 = πR^2/16.

Hence, we get:

Φ = q/(32ε0L) R^2

Conclusion

The amount of flux passing through the disc is given by Φ = q/(32ε0L) R^2. This result shows that the