A very long conducting cylindrical rod of length l if the total charge is -2q . use gauss's law to find electrical field at

a} r

b}a

c}r>b

To find the electric field around a long conducting cylindrical rod with a total charge of "-2q" using Gauss's Law, we need to consider the geometry of the situation and apply the law appropriately in different regions around the rod. Gauss's Law states that the electric flux through a closed surface is proportional to the enclosed charge. The mathematical expression for Gauss's Law is given by:

Φ = ∮ E · dA = Q_enc / ε₀

Where:

• Φ is the electric flux through the surface.
• E is the electric field.
• dA is the differential area vector.
• Q_enc is the charge enclosed by the surface.
• ε₀ is the permittivity of free space.

Case a: r < a

In this scenario, we are looking at a point inside the conducting rod. Since the rod is a conductor, the electric field inside a conductor in electrostatic equilibrium is zero. Therefore, for any point where the distance from the center of the rod (r) is less than the radius of the rod (a), the electric field (E) is:

E = 0

Case b: a < r < b

Now, we consider a point outside the conducting rod but still within the cylindrical shell formed by the rod. To apply Gauss's Law, we choose a cylindrical Gaussian surface of radius r (where a < r < b) and length l. The charge enclosed by this Gaussian surface is still the total charge of the rod, which is "-2q". The electric field is uniform and directed radially outward (or inward, depending on the sign of the charge) over the curved surface of the cylinder.

Using Gauss's Law:

Φ = E(2πrl) = Q_enc / ε₀

Substituting the enclosed charge:

E(2πrl) = -2q / ε₀

Solving for E gives:

E = -2q / (2πr l ε₀)

This simplifies to:

E = -q / (πr l ε₀)

Case c: r > b

In this case, we are looking at a point outside the cylindrical rod. Again, we will use a cylindrical Gaussian surface of radius r (where r > b). The charge enclosed by this surface remains the same, which is "-2q". The application of Gauss's Law is similar to the previous case:

Φ = E(2πrl) = Q_enc / ε₀

Substituting the enclosed charge:

E(2πrl) = -2q / ε₀

Solving for E gives:

E = -2q / (2πrl ε₀)

This simplifies to:

E = -q / (πrl ε₀)

Summary of Results

To summarize the electric field in different regions around the conducting cylindrical rod:

• For r < a: E = 0
• For a < r < b: E = -q / (πrl ε₀)
• For r > b: E = -q / (πrl ε₀)

This analysis shows how the electric field behaves in relation to the charge distribution on a conducting cylindrical rod, illustrating the principles of electrostatics and Gauss's Law effectively.