A very long cylindrical shell with a radius of 50 mm carries a uniform surface charge density of \( 1.0 \times 10^{-9} \mathrm{C} / \mathrm{m}^{2} \). What is the electrig-field strength at a distance of 2.7 mm from the surface of the shell.

To find the electric field strength at a distance of 2.7 mm from the surface of a long cylindrical shell with a uniform surface charge density, we can use Gauss's law. The total distance from the center of the cylinder to the point where you want to find the electric field is \( r = 50 \, \text{mm} + 2.7 \, \text{mm} = 52.7 \, \text{mm} \). Using Gauss's law, the electric field \( E \) at that distance from the surface of a charged cylinder is given by: \[ E = \frac{\sigma}{2\pi \epsilon_0 r} \] Where: - \( \sigma = 1.0 \times 10^{-9} \, \text{C/m}^2 \) (surface charge density) - \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{(N m}^2\text{)} \) (permittivity of free space) - \( r = 52.7 \times 10^{-3} \, \text{m} \) (distance from the center). Plugging in the values: \[ E = \frac{1.0 \times 10^{-9}}{2\pi (8.85 \times 10^{-12})(52.7 \times 10^{-3})} \] Calculating this gives \( E \approx 34.5 \, \text{N/C} \). Therefore, the electric field strength at a distance of 2.7 mm from the surface of the shell is approximately 34.5 N/C.