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.

The force on a charge in a static electric field is not determined by potential alone; it depends on the electric field, which is not given.

Work done W = q * ΔV = 3 × 10^-6 C * (600 V - 300 V) = 3 × 10^-6 * 300 = 0.9 mJ.

Distance d = V/E = 50 V / 5 N/C = 10 m.

Change in potential energy ΔU = qΔV = 3 C × (15 V - 5 V) = 30 J.

An increase in electric potential at a point generally indicates a stronger electric field, as the electric field is related to the rate of change of potential.

If the electric potential increases, the work done by the external force must increase to move the positive charge against the electric field.

The work done (W) by the electric field is given by W = q(V_A - V_B). For a unit charge, W = 10 V - 5 V = 5 J.

Potential difference (V_AB) = V_A - V_B = 15 V - 5 V = 10 V.

The potential difference V_AB = V_B - V_A = 15 V - 5 V = 10 V.

Work done W = Q(V_B - V_A) = 2 C * (15 V - 5 V) = 20 J.

If the electric potential is constant, the electric field in that region is zero, as E = -dV/dx.

The electric field at the surface of a sphere is given by E = Q/(4πε₀R²). If R is halved, E increases by a factor of 4.

The electric field E due to a point charge decreases with the square of the distance from the charge, so if the radius is doubled, the electric field halves.

The electric field due to a point charge is independent of the radius of the Gaussian surface; it remains the same.

The electric field E due to a point charge decreases with the square of the distance from the charge, so if the radius is doubled, the electric field halves.

If the total charge enclosed is zero, the electric field can still be non-zero at points on the surface due to external charges.