Coulomb's Law (force between multiple charges)

Coulomb’s law states that the force magnitude is proportional to |q₁q₂|, while the signs determine whether the force is attractive or repulsive. Initially, attraction means the charges have opposite signs. Reversing the sign of one charge makes their signs the same, so the interaction changes to repulsion. Increasing only one charge magnitude by a factor of 15 multiplies |q₁q₂|, and hence the force magnitude, by 15. Therefore the new force is 15 times repulsive.

Repulsion initially means that the two charges have the same sign. Reversing the sign of only one makes their signs opposite, so the force changes from repulsive to attractive. The magnitude of each charge becomes 1/8 of its original value, so the charge product becomes (1/8)(1/8) = 1/64 of its original value. Since distance is unchanged, Coulomb’s law says the force magnitude also becomes one-sixty-fourth. Thus the force is one-sixty-fourth attractive.

Let the side of the square be a. The distance from the selected corner to an adjacent corner is a, whereas the distance to the opposite corner is the diagonal, √2a. Coulomb’s law gives F ∝ 1/r² for equal charges. Therefore F_opposite/F_adjacent = a²/(√2a)² = a²/(2a²) = 1/2. The opposite-corner force is consequently half the adjacent-corner force. The charge values cancel in the comparison.

The centre is equidistant from all four corners, and every corner charge has the same magnitude. Consider one pair of opposite corners: the two forces on the target have equal magnitudes because the distances and charge magnitudes are equal, but their directions are opposite along the same diagonal, so they cancel. The other opposite pair cancels in exactly the same way, regardless of the target charge sign. Therefore the total electrostatic force is zero.

The governing idea is electrostatic induction and the resulting unequal electric forces. A neutral conductor contains both positive nuclei and mobile electrons, with equal total amounts only in the net-charge sense. A nearby charged object separates these charges, placing opposite charge closer to itself and like charge farther away. The closer opposite-charge attraction is stronger, so the net force is attractive. Therefore A is correct; neutrality does not mean absence of charges.

Metals contain a lattice of positive ions whose nuclei remain bound to fixed positions, while some outer electrons are delocalized and can move through the material. When a potential difference is applied, these free electrons drift and produce charge transfer or current. Protons, neutrons, and complete nuclei cannot freely travel through the metallic lattice. Therefore, option C is correct.

This is charging by induction, and the sequence is essential. The nearby positive rod attracts electrons toward the conductor. While the rod remains present, earthing allows additional electrons to flow from Earth into the conductor. Removing the earth connection first traps this excess negative charge; removing the rod afterward only redistributes it over the conductor. Therefore the final charge is negative, so option B is correct.

Use the vector relation F = qE. The field magnitude is |E| = |F|/|q| = 8 N/2 C = 4 N C⁻¹. Since the charge is negative, its force is opposite to the electric field. The force points west, so the electric field must point east. Thus option B gives both the correct magnitude and direction; option A ignores the negative sign.

The net electric field at a zero-field point is E_net = 0, meaning the vector contributions from the two positive charges cancel exactly at that location. The force on a test charge is F = qE_net. Even though each positive charge exerts an individual force, their vector sum is zero, so the net force on the small positive test charge is zero. Thus option A is correct; the size-based directions in B and C ignore cancellation.

The governing relation is vector force on a charge: F = qE. The magnitude is |F| = |q|E = 3 × 12 = 36 N. Because q is negative, the force direction is opposite to the upward electric field, namely downward. Therefore option B is correct. Option A has the right magnitude but wrong direction, while C and D result from incorrect arithmetic or formula use.

The positive test charge is attracted toward each negative source charge. Because the source charges are equal and the test charge is exactly at the midpoint, its distances from both sources are equal, so Coulomb’s law gives two equal force magnitudes. The forces point in opposite directions along the same line and cancel vectorially, producing zero net force. Option A is correct; attraction toward one side would require unequal conditions, while C and D have incorrect directions.

A dipole in a uniform electric field experiences equal and opposite forces on its two charges. These forces have zero net force but generally form a couple, producing torque τ = pE sin θ. The torque rotates the dipole and tends to align its dipole moment with the field, so option A is correct. It does not change the charges or their separation; at alignment, the torque becomes zero.

Electric field lines are drawn in the direction of the force on a small positive test charge. They originate at positive charges and terminate at negative charges; therefore, for a dipole, the external field lines point from +q toward −q. Option B confuses field-line direction with dipole-moment direction. Lines do not generally move outward from both charges or inward toward both charges.

The torque on an electric dipole is τ = pE sin θ, where θ is the angle between the dipole moment and the electric field. When the dipole is aligned with the field, θ = 0° and sin 0° = 0. Therefore τ = 0, so there is no turning tendency. Maximum torque occurs at 90°, not at 0°. Doubling or infinite torque has no basis when p, E, and the angle are otherwise finite. Hence option A is correct.

The torque on an electric dipole in a uniform field is τ = pE sin θ. Assuming p and E remain constant, at 30° the torque is pE sin 30° = 0.5pE. At 90°, it becomes pE sin 90° = pE, its maximum value. Since pE is twice 0.5pE, the torque increases, specifically becoming twice its initial value. Therefore option A is correct; it neither decreases nor remains unchanged.