A thin insulating rod with charge density λ=+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 electric field at point P, which is 6cm from the central axis? Give the electric field magnitude in N/C.
To find the electric field at point P outside the conducting cylindrical shell, we can use Gauss's Law. Gauss's Law states that the total electric flux through a closed surface is equal to the total charge enclosed divided by the permittivity of free space.
In this case, we can use a Gaussian cylinder with radius r = 6 cm and length L that encompasses the insulating rod. The electric field will be radially symmetric, so we can consider a cross-sectional area of the Gaussian cylinder.
The charge enclosed within the Gaussian cylinder can be found by multiplying the charge density of the rod, λ, by the length of the Gaussian cylinder, L.
q = λL
To find the total electric flux through the Gaussian cylinder, we can apply Gauss's Law:
Φ = q / ε₀
where ε₀ is the permittivity of free space. The electric field E is perpendicular to the surface of the Gaussian cylinder, so the flux is given by:
Φ = E * (2πrL)
Setting these equations equal to each other and solving for E:
E * (2πrL) = q / ε₀
E = q / (2πrLε₀)
Substituting the expression for q:
E = λL / (2πrLε₀)
Canceling out the L terms:
E = λ / (2πrε₀)
Now we can substitute in the values given in the problem:
λ = +5 nC/m = 5 * 10^(-9) C/m
r = 6 cm = 0.06 m
ε₀ = 8.85 * 10^(-12) C^2/(N*m^2)
Plugging in these values:
E = (5 * 10^(-9) C/m) / (2π * 0.06 m * 8.85 * 10^(-12) C^2/(N*m^2))
E ≈ 450 N/C
Therefore, the magnitude of the electric field at point P, which is 6 cm from the central axis, is approximately 450 N/C.