Field of uniformly charged cylinder
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Exercise:
A long hollow cylinder is uniformly charged with linear charge density lambda. abcliste abc Using Gauss's law show that there is no electric field on the inside of the cylinder. abc Using Gauss's law show that the electric field on the outside of the cylinder is given by Er fracpivarepsilon_ fraclambdar where r is the distance from the central axis of the cylinder. abcliste
Solution:
abcliste abc If there is an electric field on the inside of the cylinder it has to respect the symmetry of the charge distribution i.e. the field lines are perpicular to the inner wall of the cylinder an po towards the centre or away from it and the magnitude of the field vectors is constant for a given distance from the centre. An appropriate closed surface Gaussian surface is therefore a cylinder with radius rR where R is the radius of the cylinder. The electric flux through this cylinder is then see inner green cylinder in figure center includegraphicswidthtextwidth#image_path:gauss-cylinder-# center Phi_E Er Ar Er pi r L where L is the length of the cylinder. The flux through the circular pieces is zero since the field lines are parallel. Using Gauss's law Phi_E fracsscQenclosedvarepsilon_ it follows that Er for rR abc For the field on the outside the cylindrical symmetry again implies a radial field. In order to find an expression for the electric field at distance r' R from the cylinder axis we choose a cylindrical surface with the corresponding radius. The electric flux is see a Phi_E Er' Ar Er' pi r' L Using Gauss's law it follows that Er' fracsscQenclosedpi r' L varepsilon_ fracpivarepsilon_ fraclambdar' quad square abcliste
A long hollow cylinder is uniformly charged with linear charge density lambda. abcliste abc Using Gauss's law show that there is no electric field on the inside of the cylinder. abc Using Gauss's law show that the electric field on the outside of the cylinder is given by Er fracpivarepsilon_ fraclambdar where r is the distance from the central axis of the cylinder. abcliste
Solution:
abcliste abc If there is an electric field on the inside of the cylinder it has to respect the symmetry of the charge distribution i.e. the field lines are perpicular to the inner wall of the cylinder an po towards the centre or away from it and the magnitude of the field vectors is constant for a given distance from the centre. An appropriate closed surface Gaussian surface is therefore a cylinder with radius rR where R is the radius of the cylinder. The electric flux through this cylinder is then see inner green cylinder in figure center includegraphicswidthtextwidth#image_path:gauss-cylinder-# center Phi_E Er Ar Er pi r L where L is the length of the cylinder. The flux through the circular pieces is zero since the field lines are parallel. Using Gauss's law Phi_E fracsscQenclosedvarepsilon_ it follows that Er for rR abc For the field on the outside the cylindrical symmetry again implies a radial field. In order to find an expression for the electric field at distance r' R from the cylinder axis we choose a cylindrical surface with the corresponding radius. The electric flux is see a Phi_E Er' Ar Er' pi r' L Using Gauss's law it follows that Er' fracsscQenclosedpi r' L varepsilon_ fracpivarepsilon_ fraclambdar' quad square abcliste