Question

state Gauss law theorem and apply to find the electric field at a point due to a) line of charge b) a plane sheet of charge c) a charged spherical conducting shell.

Answer 1

Gauss's Law. The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux through an area is defined as the electric field multiplied by the area of the surface projected in a plane perpendicular to the field.

Answer 2

According to Gauss's law, the net electric charge enclosed by any closed surface has a direct relationship to the electric flux through that surface. It can be written mathematically as =E.A is equal to q/0, where q is the charge contained by the surface, A is the area of the closed surface, E is the electric field, and A is the permittivity of free space

a) Electric field at a point due to a line of charge:

• Consider an infinitely long straight line of charge with uniform linear charge density λ.
• To find the electric field at a point P at a distance r from the line of charge, take a cylindrical Gaussian surface of radius r and length l with its axis passing through point P and perpendicular to the line of charge.
• By Gauss's law, the electric flux through the closed surface is Φ=E.A=q/ε0 = (λl)/ε0, where A=2πrl is the area of the cylindrical surface.
• Due to symmetry, the electric field E is constant and perpendicular to the surface, and its magnitude can be found as E = (λ/(2πε0r))

b) Electric field at a point due to a plane sheet of charge:

• Consider an infinite plane sheet of charge with surface charge density σ.
• To find the electric field at a point P at a distance r from the plane, take a Gaussian cylinder of radius r and length L with its axis perpendicular to the plane.
• By Gauss's law, the electric flux through the closed surface is Φ=E.A=q/ε0 = σA/ε0, where A=2πrL is the area of the curved surface of the cylinder.
• Due to symmetry, the electric field E is constant and perpendicular to the surface, and its magnitude can be found as E = σ/(2ε0)

c) Electric field at a point due to a charged spherical conducting shell:

• Consider a spherical conducting shell of radius R with total charge Q distributed uniformly on its surface.
• To find the electric field at a point P inside or outside the shell, take a spherical Gaussian surface of radius r with its center at the center of the shell.
• By Gauss's law, the electric flux through the closed surface is Φ=E.A=q/ε0 = Q/ε0, where A=4πr^2 is the area of the spherical surface.
• Due to symmetry, the electric field E is constant and radial, and its magnitude can be found as E = Q/(4πε0r^2) for r>R (outside the shell), and E=0 for r<R (inside the shell).

To learn more about Gauss's law from the given link.

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