Positive charge is distributed throughout a long non-conducting cylindrical shell of inner
radius R and outer radius 2R. At what radial depth beneath the outer surface of the charge
distribution is the electric field strength equal to one-half the surface value?
To determine the radial depth beneath the outer surface of the charge distribution where the electric field strength is equal to one-half the surface value, we can use Gauss's law.
Gauss's law states that the electric flux through a closed surface is equal to the net charge enclosed divided by the permittivity of the medium.
In this case, the electric field inside the cylindrical shell is constant and given by:
E_surface = σ / (2ε₀)
where σ is the surface charge density, and ε₀ is the permittivity of free space.
To find the depth beneath the outer surface where the electric field strength is half the surface value, we need to find the distance from the outer surface (2R) to that depth.
Let's assume the depth we are looking for is x. The electric field at this depth is E_depth = 0.5 * E_surface.
Now, we need to find the charge enclosed within a cylindrical Gaussian surface of radius x.
The charge inside the Gaussian surface can be calculated as:
Q_enclosed = σ * (Area of the Gaussian surface)
The area of the Gaussian surface is given by:
A = 2πxL
where L is the length of the cylindrical shell.
Using Gauss's law, we can equate the electric flux through the Gaussian surface to the charge enclosed:
E_depth * (Area of the Gaussian surface) = Q_enclosed / ε₀
(0.5 * E_surface) * (2πxL) = (σ * 2πxL) / ε₀
Simplifying, we find:
0.5 * (σ / (2ε₀)) * (2πxL) = (σ * 2πxL) / ε₀
Simplifying further, we get:
x = R / 2
Therefore, the radial depth beneath the outer surface of the charge distribution where the electric field strength is equal to one-half the surface value is R/2.