For points outside the sphere, the electric field behaves as if all the charge were concentrated at the center, so E = Q/4πε₀r².

The total flux through the cube is Q/ε₀, and since there are 6 faces, the flux through one face is Q/6ε₀.

The total flux through the cube is Q/ε₀, and since the charge is at one corner, the flux through one face is Q/(12ε₀).

The charge at one corner contributes 1/8 of its flux to the cube. Thus, total flux = (1/8)Q/ε₀ = Q/(12ε₀).

For a point outside the cylinder, the electric field is given by E = λ/(2πε₀r).

The electric flux through the curved surface is proportional to the charge enclosed, which remains constant, so the flux through the curved surface doubles if the height is doubled.

The electric flux through the curved surface of a cylinder is given by Φ = Q_enc/ε₀, where Q_enc = Q.

The electric field due to a uniformly charged infinite cylinder depends only on the charge per unit length, not on the radius.

Gauss's law applies to electric fields, not magnetic fields. The electric field around a current-carrying wire is not defined by Gauss's law.

The electric field outside the cylindrical surface is directly proportional to the distance from the wire.

Using Gauss's law, the electric field outside the cylinder is E = Q/(2πε₀R).

Using Gauss's law, the electric field E at a distance R from the axis of a long charged cylinder is E = Q/(2πε₀L) for points outside the cylinder.

For a hollow cylinder, the electric field at a point outside is E = λ/(2πε₀r) using Gauss's law.

The electric field inside a hollow charged sphere is zero due to symmetry.

Inside a hollow charged sphere, the electric field is zero due to symmetry.

The electric field inside a hollow charged sphere is zero due to symmetry.

Inside a hollow charged sphere, the electric field is zero due to symmetry, as per Gauss's law.

Inside the cavity of a hollow conductor, the electric field is zero due to the shielding effect of the conductor.

Using Gauss's law for a cylindrical surface around the wire, the electric field E at a distance r is E = λ/(2πε₀r).

The electric flux through the Gaussian surface is given by Φ = Q/ε₀, where Q is the charge enclosed.

The total electric flux through the surface is given by Gauss's law as Φ = Q/ε₀, and for a point charge at the center, it results in 4πQ/ε₀.

According to Gauss's law, the total electric flux through a closed surface is Φ = Q_enc/ε₀. Here, Q_enc = Q, so Φ = Q/ε₀.

According to Gauss's law, the electric field inside a conductor in electrostatic equilibrium is zero.

The total flux through the cube is Q/ε₀. Since there are 6 faces, the flux through one face is Q/(6ε₀).

According to Gauss's law, the electric flux Φ = Q/ε₀ when a point charge Q is at the center of a spherical surface.

The electric field outside a spherical charge distribution is given by E = Q/4πε₀r². At 2R, it becomes Q/4πε₀(2R)².

For a spherical shell, the electric field outside the shell behaves as if all the charge were concentrated at the center, so E = Q/(4πε₀R²).

By Gauss's law, the electric field inside a uniformly charged spherical shell is zero.

By Gauss's law, the electric field inside a uniformly charged spherical shell is zero.

For a uniformly charged sphere, the electric field outside the sphere behaves as if all the charge were concentrated at the center, thus E = Q/(4πε₀r²).