The gravitational potential is inversely proportional to the distance, so if the distance is doubled, the potential halves.

The gravitational field strength varies inversely with the square of the distance, so if the distance is doubled, the field strength becomes 1/4.

The gravitational field strength varies inversely with the square of the distance from the center of the Earth, so if the distance is doubled, the field strength becomes one-fourth.

If the Earth shrinks in size while maintaining its mass, the gravitational force at the surface would increase due to the decrease in radius.

If the radius is halved, the gravitational force increases by a factor of 4, since F = GM/R^2.

If the radius is halved, the gravitational acceleration would increase by a factor of 4 (g ∝ 1/R²).

The gravitational force at the surface is inversely proportional to the square of the radius. If the radius is halved, the force becomes four times stronger.

Force F = E * q = 100 N/C * 5 × 10^-6 C = 0.0005 N = 0.5 N.

F = E * q = 200 N/C * 5 × 10^-6 C = 1 N.

The electric field is directed towards the charge, indicating that the charge is negative.

V = -E * d = -200 N/C * 3m = -600 V. The potential at B is 600 V lower than at A.

The electric field is zero when the vector sum of the electric fields due to all charges is zero, which occurs when there are equal and opposite charges.

The electric field due to an infinite plane sheet is constant and equal to E on both sides, thus the total field is 2E.

The electric field due to an infinite plane sheet is constant and equal to E, regardless of the distance from the sheet.

The fields due to the two sheets add up in the region between them, resulting in a total electric field of 2E.

Using E = k * |q| / r^2, we find q = E * r^2 / k = 1000 * 1^2 / (9 × 10^9) = 1μC.

The electric field due to a point charge varies inversely with the square of the distance, so at 2r, E becomes E/4.

In electrostatic equilibrium, excess charge resides on the surface of the conductor, leading to zero electric field inside.

The electric field just outside a conductor in electrostatic equilibrium is perpendicular to the surface.

The electric field is the negative gradient of the electric potential. If the potential is constant (0 V), the electric field is zero.

Electric potential energy U is given by U = VQ. Here, U = 10 V * 2 C = 20 J.

Work done W = q(V1 - V2) = 2 C (10 V - 0 V) = 20 J.

Work done = Charge × Potential = 2 C × 10 V = 20 J.

Distance d = V / E = 100 V / 50 N/C = 2 m.

Work done W = q * V = 2 C * 100 V = 200 J.

If the electric field is directed towards the point, it indicates that the charge creating the field is negative.

Potential difference ΔV = E * d = (150 V / 3 m) * 3 m = 150 V.

Potential energy U = q * V = -1 C * 200 V = -200 J.

Work done W = q * V = 0.5 C * 200 V = 100 J.

Work done W = q * (V1 - V2) = 0.5 C * (200 V - 100 V) = 50 J.