Spherical Capacitor
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Exercise:
A spherical capacitor consists of two concentric conducting shells with charges pm Q and radius r_i inner shell and r_o outer shell respectively. abcliste abc Show that there is no electric field on the inside of the inner shell r r_i. abc Derive an expression for the electric field between the two shells r_i r r_o. abc Show that there is no electric field outside the outer shell r r_o. abcliste
Solution:
In each case the symmetry of the field radial field lines implies that a concentric shell is an appropriate Gaussian surface. Asing the electric field at distance r from the centre is Er the electric flux is Phi_E Er Ar pi Er r^ abcliste abc There is no charge inside the inner shell so for r r_i Gauss's law yields Phi_E pi Er r^ fracsscQenclosedvarepsilon_ It follows that Er for r r_i. abc The charge enclosed by a Gaussian surface with r_i r r_o is the charge on the inner shell i.e. Q. Using Gauss's law Phi_E pi Er r^ fracsscQenclosedvarepsilon_ fracQvarepsilon_ it follows Er fracQpivarepsilon_ r^ abc For a closed surface containing both shells the enclosed charge is zero. Using Gauss's law Phi_E pi Er r^ fracsscQenclosedvarepsilon_ which yields Er for r r_o. abcliste The figure below displays the electric as a function of r. center includegraphicswidthtextwidth#image_path:spherical-capacitor-# center
A spherical capacitor consists of two concentric conducting shells with charges pm Q and radius r_i inner shell and r_o outer shell respectively. abcliste abc Show that there is no electric field on the inside of the inner shell r r_i. abc Derive an expression for the electric field between the two shells r_i r r_o. abc Show that there is no electric field outside the outer shell r r_o. abcliste
Solution:
In each case the symmetry of the field radial field lines implies that a concentric shell is an appropriate Gaussian surface. Asing the electric field at distance r from the centre is Er the electric flux is Phi_E Er Ar pi Er r^ abcliste abc There is no charge inside the inner shell so for r r_i Gauss's law yields Phi_E pi Er r^ fracsscQenclosedvarepsilon_ It follows that Er for r r_i. abc The charge enclosed by a Gaussian surface with r_i r r_o is the charge on the inner shell i.e. Q. Using Gauss's law Phi_E pi Er r^ fracsscQenclosedvarepsilon_ fracQvarepsilon_ it follows Er fracQpivarepsilon_ r^ abc For a closed surface containing both shells the enclosed charge is zero. Using Gauss's law Phi_E pi Er r^ fracsscQenclosedvarepsilon_ which yields Er for r r_o. abcliste The figure below displays the electric as a function of r. center includegraphicswidthtextwidth#image_path:spherical-capacitor-# center