A thin insulating rod with charge density +5nC/m is arranged inside a thin conducting cylindrical shell of radius R=3cm. The rod and shell are on the same axis, and you can assume they are both infinite in length. What is the surface charge density induced on the outside of the conducting shell in nC/m^2?

I thought since the rod and shell is infinite in length the electric field is zero inside. Therefore, the charge density on the outside of the conducting shell is - -5nC/m. That seemed too easy.

Then my next answer was
charge density(outer) =
-(lambda/2*pi*r)= - (5/(2*pi*0.03) = -26.5258 nC/m^2

I have no other solutions.

Thank you.

Only it says on the outside of the conducting shell. Be sure to check the positive/negative signs in your answer.

Your initial intuition is correct that the electric field inside the conductor is zero. However, the surface charge density induced on the outside of the conducting shell is not simply the opposite of the charge density of the rod.

To determine the surface charge density induced on the outside of the conducting shell, we need to consider the electric field just outside the rod. Let's calculate this electric field using Gauss's law.

First, let's consider a Gaussian surface in the form of a cylinder with radius r and length L. The Gaussian surface encloses both the rod and the conducting shell.

According to Gauss's law, the electric flux through the Gaussian surface is equal to the charge enclosed by the surface divided by the permittivity of free space.

Since the electric field inside the conductor is zero, the only charge enclosed by the Gaussian surface is the charge from the rod.

Therefore, the electric flux through the Gaussian surface is:

Flux = (Charge enclosed) / (permittivity of free space)

Using Gauss's law, we can write:

Electric field just outside the rod * (2πrL) = (Charge enclosed) / (permittivity of free space)

The charge enclosed by the Gaussian surface is equal to the charge density of the rod multiplied by the length L:

Charge enclosed = (Charge density of the rod) * L

Rewriting the equation using these values, we get:

Electric field just outside the rod * (2πrL) = (Charge density of the rod) * L / (permittivity of free space)

Simplifying, we find:

Electric field just outside the rod = (Charge density of the rod) / (2πε₀r)

Now, we can use this expression for the electric field just outside the rod to find the induced surface charge density on the conducting shell.

For a conductor in electrostatic equilibrium, the electric field just outside the conductor is perpendicular to its surface. Therefore, the surface charge density induced on the conductor is given by:

Surface charge density induced on the conductor = ε₀ * (Electric field just outside the rod)

Substituting the expression for the electric field just outside the rod, we get:

Surface charge density induced on the conductor = ε₀ * (Charge density of the rod) / (2πr)

Plugging in the given values of the charge density of the rod (+5nC/m) and the radius of the conducting shell (0.03m), and using the value of ε₀, the permittivity of free space, we find:

Surface charge density induced on the conductor = (8.854 x 10^-12 C²/Nm²) * (5 x 10^-9 C/m) / (2π x 0.03m)

Surface charge density induced on the conductor ≈ 7.429 x 10^-6 C/m²

Converting to nC/m², we find:

Surface charge density induced on the conductor ≈ 7.429 nC/m²

Therefore, the correct surface charge density induced on the outside of the conducting shell is approximately 7.429 nC/m².

To find the surface charge density induced on the outside of the conducting shell, you need to consider the electric field produced by the charged rod inside the shell.

The electric field at any point outside the charged rod can be found using Gauss's law. Since the rod and shell are both infinite in length and symmetrically arranged, we can use a cylindrical Gaussian surface just outside the rod, with radius R.

Since the electric field inside the conducting shell is zero, the electric field on the Gaussian surface is also zero. This means that the net charge enclosed by the Gaussian surface must be zero.

The linear charge density of the rod is +5nC/m, which means that the charge enclosed by the Gaussian surface is λ * 2πr, where λ is the charge density of the rod and r is the radius of the Gaussian surface.

Setting the net charge enclosed by the Gaussian surface to zero, we can solve for the surface charge density induced on the outside of the conducting shell:

λ * 2πr = σ * 2πr²

where σ is the surface charge density on the outside of the shell and r is the radius of the Gaussian surface (which is equal to the radius of the shell, R).

Plugging in the values, we get:

+5nC/m * 2πR = σ * 2πR²

Simplifying the equation, we find:

σ = +5nC/m * R/ R²

σ = +5nC/m * (1/R)

Now, you can substitute the given value of R (0.03m) to find the surface charge density induced on the outside of the conducting shell:

σ = +5nC/m * (1/0.03)

σ = +166.67 nC/m²

Therefore, the surface charge density induced on the outside of the conducting shell is +166.67 nC/m².