V = k * q / r = (9 × 10^9 N m²/C²) * (3 × 10^-6 C) / 0.3 m = 9000 V.

Work done W = F * d = (E * q) * d = (1500 N/C * 3 × 10^-6 C) * 0.2 m = 0.0009 J = 90 J.

E = k * |q| / r^2 = (9 × 10^9) * 4 × 10^-6 / (2)^2 = 450 N/C.

E = k * |q| / r² = (9 × 10^9 N m²/C²) * (5 × 10^-6 C) / (2 m)² = 1125 N/C.

Force F = qE = (5 × 10^-6 C) * (2000 N/C) = 10 N.

Potential energy U = q * V = -2 × 10^-6 C * 1000 N/C = -0.002 J.

Force F = E * q = 500 N/C * -2 × 10^-6 C = -1 N.

Potential energy U = q * V = -4 × 10^-6 C * 200 N/C = -800 μJ.

Distance r = √(3² + 4²) = 5 m. V = k * q / r = (9 × 10^9) * (4 × 10^-6) / 5 = 720 V.

V = k * q / r = (9 × 10^9 N m²/C²) * (4 × 10^-6 C) / (3 m) = 1200 V.

Potential energy U = q * V, where V = E * d. Assuming d = 1 m, U = 4 × 10^-6 C * 1000 N/C * 1 m = 4 mJ.

Electric potential energy (U) = Charge × Electric Field = 5 μC × 2000 N/C = 10 mJ.

Potential energy (U) = Charge × Electric Field × Distance. Assuming distance = 1 m, U = 5 μC × 2000 N/C = 10 mJ.

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

Charge Q = C × V = 4 µF × 12 V = 48 µC.

As the charged particle moves to a lower potential, it loses potential energy, which is converted into kinetic energy, thus increasing its kinetic energy.

As the charged particle moves from high potential to low potential, it loses potential energy, which is converted into kinetic energy, thus its kinetic energy increases.

In a uniform electric field, the force acting on a charged particle is constant in magnitude and direction.

For a charged sphere, the electric potential outside the sphere behaves as if all the charge were concentrated at the center, so V = kQ/r.

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

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

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

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.

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.