Force F = qE = (1 × 10^-6 C)(1000 N/C) = 0.001 N.

Force F = qE = (10 × 10^-6 C) * (2000 N/C) = 0.02 N = 20 N.

Force F = qE = (10 × 10^-6 C) * (2000 N/C) = 0.02 N = 20 N.

Work done W = F * d = (E * q) * d = (1000 N/C * 5 × 10^-6 C) * 0.2 m = 0.1 J.

Work done W = F * d = (E * q) * d = (300 N/C * 5 × 10^-6 C) * 0.2 m = 0.3 J.

The force on a negative charge is in the direction opposite to the electric field, hence towards the positive charge.

Force F = qE = (1 × 10^-6 C) * (1000 N/C) = 0.001 N.

Force F = qE = (1 × 10^-6 C) * (2000 N/C) = 0.002 N.

Force F = qE = (10 × 10^-6 C) * (200 N/C) = 2 N.

Capacitance C = Q/V. If Q is doubled and V remains constant, C must also double.

The electric field E between the plates of a capacitor is directly proportional to the charge. If the charge is doubled, the electric field also doubles.

The electric field E is directly proportional to the charge, so tripling the charge will triple the electric field.

The electric fields due to both charges at the midpoint cancel each other out, resulting in a net electric field of 0 N/C.

The electric fields due to both charges at the midpoint will cancel each other, resulting in a net electric field of 0 N/C.

E = V/d = 10 V / 2 m = 5 N/C.

The electric field E due to a dipole at a point along the axial line is given by E = (1/4πε₀) * (2p) / r³.

According to Coulomb's law, F ∝ 1/r². Reducing the distance to half increases the force by a factor of 4.

According to Coulomb's law, F ∝ q1 * q2. If both charges are doubled, the force becomes 4 times the original force.

The electric field E is directly proportional to the charge, so if the charge is doubled, the electric field also doubles.

E = k * |q| / r² = (9 × 10^9 N m²/C²) * (10 × 10^-6 C) / (1 m)² = 9000 N/C.

E = k * |q| / r² = (9 × 10^9 N m²/C²) * (4 × 10^-6 C) / (1 m)² = 3600 N/C, directed towards the charge.

The electric field due to an infinite plane sheet is E = σ/2ε₀ on both sides, but the total field is σ/ε₀.

Electric field E = k * |q| / r² = (9 × 10^9 N m²/C²) * (10 × 10^-6 C) / (0.1 m)² = 9000 N/C.

Using E = k * |q| / r² = (9 × 10^9 N m²/C²) * (5 × 10^-6 C) / (0.2 m)² = 11250 N/C.

Potential difference V = E * d = 50 N/C * 0.4 m = 20 V.

Potential difference V = E * d = 400 N/C * 0.5 m = 200 V.

Potential energy U = k * q1 * q2 / r = (9 × 10^9 N m²/C²) * (3 × 10^-6 C) * (4 × 10^-6 C) / 0.2 m = 54 J.

Potential energy U = k * q1 * q2 / r = (9 × 10^9 N m²/C²) * (3 × 10^-6 C) * (4 × 10^-6 C) / 0.3 m = 0.36 J.

Potential energy U = k * q1 * q2 / r = (9 × 10^9 N m²/C²) * (3 × 10^-6 C) * (4 × 10^-6 C) / 0.2 m = 54 J.

Electric field lines cannot intersect because it would imply two different directions of the electric field at a point.