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 potential difference (V) between the plates is given by V = Ed, where E is the electric field and d is the distance between the plates.

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

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 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.

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.