Charge is distributed through an infinitely long cylinder of radius R in such a way that the charge density is proportional to the distance from the central axis: ß = A r, where A is a constant and ß is the density.
(a) Calculate the total charge contained in a segment of the cylinder of length L.
(b) Calculate the electric field for points outside the cylinder.
(c) Calculate the electric field for points inside the cylinder.
total charge=INTEGRAL sigma dV=INT L*Ardr
=AR^2 L
To solve this problem, you will need to use calculus and integration. Let's break down each part of the problem:
(a) To calculate the total charge contained in a segment of the cylinder of length L, we need to integrate the charge density (ß = A r) over the length L. The charge dq in an infinitesimally small segment dl of the cylinder is given by dq = ß dl = A r dl.
To find the total charge, we integrate dq with respect to dl from 0 to L:
Q = ∫ dq = ∫ A r dl
Since ß = A r, we can substitute ß for A r in the integral:
Q = ∫ ß dl = ∫ ß₀ l dl = ß₀ ∫ l dl
Integrating l with respect to l gives:
Q = ß₀ ∫ l dl = ß₀ (l² / 2)
So, the total charge contained in a segment of length L is Q = ß₀ (1/2) L².
(b) To calculate the electric field outside the cylinder, we can use Gauss's Law. Since the cylinder has infinite length, we can imagine it as a long tube of charge. The electric field outside the cylinder can be obtained by considering a Gaussian surface in the form of a cylindrical surface with radius r > R and applying Gauss's Law:
∮ E · dA = Q_enclosed / ε₀
The electric field E is perpendicular to the cylindrical surface, so E · dA = E dA. The flux through the curved part of the Gaussian surface is zero, so only the top and bottom circular faces contribute to the flux.
The charge enclosed by the Gaussian surface is the total charge of the cylinder. So Q_enclosed = ß₀ (1/2) L².
The surface area of each circular face is A = πr². Therefore, the total surface area of both faces is 2πr².
Applying Gauss's Law, we have:
∮ E · dA = E ∮ dA = E * 2πr² = Q_enclosed / ε₀
Solving for E, we get:
E = Q_enclosed / (2πr²ε₀)
Substituting Q_enclosed = ß₀ (1/2) L², we get:
E = (ß₀ (1/2) L²) / (2πr²ε₀)
(c) To calculate the electric field inside the cylinder, we will again use Gauss's Law. This time, we consider a Gaussian surface in the form of a cylindrical surface with radius r < R.
Since the charge inside the Gaussian surface is zero (all charge is located outside), the total flux through the surface is zero. Therefore, the electric field inside the cylinder is zero.
So, the electric field for points inside the cylinder is E = 0.