Coulomb's Law (force between multiple charges)

Coulomb’s law gives F = k|q1q2|/r². Since the distance and medium do not change, only the charge product changes. The new product is (q1/2)(q2/3) = q1q2/6, so the new force is one-sixth of the original force. Thus F' = 48/6 = 8 N. Sixteen N would represent a one-third factor, while 24 N represents one-half, so option A is correct.

For two point charges at fixed distance, Coulomb’s law gives F ∝ |q1q2|. If both charges are doubled, their product changes from q1q2 to (2q1)(2q2) = 4q1q2. The force therefore becomes four times its original value. Halving one charge or both charges decreases the force, and doubling the distance reduces it by a factor of four rather than increasing it.

With unchanged charges, Coulomb’s law gives F ∝ 1/r². Let the new distance be r'. To obtain F' = F/4, we require 1/(r')² = (1/4)(1/r²), which gives (r')² = 4r² and hence r' = 2r for a positive distance. Therefore the separation must be doubled. Quadrupling the distance would reduce the force to one-sixteenth.

For fixed charges, Coulomb’s law states F ∝ 1/r². If the distance is changed to r/3, then the inverse-square factor becomes 1/(r/3)² = 9/r². Consequently, the force becomes 9F. Tripling the distance would instead make the force F/9, and making the distance nine times larger would reduce it even more. Hence the required change is to make the distance one-third.

The resultant of two equal forces F making an angle θ is R = √(F² + F² + 2F²cosθ). For θ = 60°, R = √(2F² + F²) = √3F, which is nonzero for any nonzero F. Two equal vectors cancel only when they point in exactly opposite directions, meaning their angle is 180°, not 60°. Therefore option B is correct.

Like charges repel. Thus each of the two equal positive source charges pushes the positive target away from itself. Because both sources lie on the same side of the target, both repulsive forces point in the same direction, away from that side. Equal vectors in the same direction add, giving a resultant whose magnitude is the sum of their magnitudes. Cancellation would require opposite directions, as in a symmetric placement on opposite sides.

Opposite charges attract, so each negative source charge pulls the positive target toward itself. Since both negative charges are on the same side of the target and are equally placed, both attractive forces point toward that same side. Their vectors therefore have the same direction and add rather than cancel. The force is not automatically zero, opposite, or perpendicular; its exact magnitude would depend on the charge values and distances.

For equal external charges placed symmetrically at equal distances, the electric forces on the target have equal magnitudes. Changing the target’s sign changes attraction into repulsion, or vice versa, so both force directions reverse simultaneously. They remain opposite to each other and therefore can still cancel. The zero resultant depends on symmetry, not on whether the target is positive or negative. Thus the condition may remain valid.

In a material medium, Coulomb’s law is modified by the medium’s permittivity; for a uniform dielectric, the electrostatic force is reduced by the relative permittivity factor compared with vacuum. In a multiple-charge problem, calculate each relevant pairwise force using the same appropriate medium factor, then add the forces as vectors. It is incorrect to ignore the medium, set every force to zero, or discard direction.

The governing concept is translational equilibrium: the resultant force on the target charge must be zero, written as ΣF = 0. This does not require every individual Coulomb force to vanish; two or more nonzero forces can cancel when their magnitudes and directions balance. Therefore option A is correct. Option B is unnecessarily restrictive, while the signs of all charges or infinite separation are not required for equilibrium.

Coulomb force is a vector, so a reliable solution begins by identifying the target charge and drawing the force due to every other charge. Calculate each magnitude from F = k|q1q2|/r², assign its direction from attraction or repulsion, and then add components or vectors. Thus option A gives the safe sequence. Options B and C ignore direction, and D changes the physical problem.

Coulomb’s law gives F ∝ q1q2/r². Doubling one charge multiplies the force by 2, whereas tripling the separation multiplies it by 1/3² = 1/9. Hence F' = 18 × 2/9 = 4 N. Option B is therefore correct. A would miss the charge factor, C would use an incorrect distance factor, and D would ignore the inverse-square dependence.

Use Coulomb’s proportionality F ∝ q1q2/r². Halving both charges changes their product by (1/2)(1/2) = 1/4. Tripling the distance contributes another factor 1/9. Therefore F' = 54 × 1/4 × 1/9 = 54/36 = 1.5 N. Option A is correct; the larger choices result from omitting one reduction or mishandling the distance square.

The governing relation is F = k(q1q2)/r². A fourfold increase in the charge product would by itself make the force four times larger. However, doubling the distance makes r² four times larger, reducing the force by a factor of four. The factors cancel: F' / F = 4/2² = 1, so option A is correct. The other options consider only one change or combine the factors incorrectly.

According to Coulomb’s law, F ∝ q1q2/r². Halving both charges changes q1q2 to one-fourth of its original value. Halving the distance changes 1/r² by a factor of four. The net multiplier is (1/4) × 4 = 1, so the force remains unchanged and option A is correct. Options C and D overlook the reduction in the charge product.

Coulomb’s law requires the factor q1q2/r². Doubling one charge gives a factor of 2. Increasing the distance fourfold makes the denominator 4² = 16 times larger, giving a factor of 1/16. Thus F' = 64 × 2/16 = 8 N, so option B is correct. Four N would result from using an incorrect distance factor, while larger options ignore the inverse-square reduction.

For unchanged charges, Coulomb’s law states F ∝ 1/r². If the new distance is r/5, then the inverse-square factor becomes 1/(r/5)² = 25/r², so the force is multiplied by 25. Therefore F' = 25 × 25 = 625 N, making option C correct. Option A uses a linear rather than square dependence, and D predicts the wrong trend.

Net force is the vector sum, so opposite-directed forces must be assigned opposite signs. Taking right as positive gives Fnet = +18 N − 11 N = +7 N. The positive sign means the resultant points to the right. Therefore option A is correct. Adding the magnitudes to obtain 29 N would be appropriate only for forces in the same direction; zero is also unjustified.

Because the forces act in opposite directions, the resultant is found by subtraction, not by adding their magnitudes. Taking left as positive, Fnet = +25 N − 9 N = +16 N. The positive sign indicates a leftward resultant, so option A is correct. Option C incorrectly adds the forces, option B reverses the direction, and option D would require equal opposing forces.

For two vectors of equal magnitude F separated by angle θ, the resultant is R = √(F² + F² + 2F² cosθ). With θ = 120°, cos120° = −1/2, so R = √(2F² − F²) = F. Therefore the resultant has the same magnitude as either individual force, making option A correct. It is not zero, because complete cancellation requires opposite forces at 180°.

The direction of a vector sum is determined by its components. Since the two forces have equal magnitudes and are symmetrically placed about the angle bisector, their components perpendicular to that bisector cancel, while the components along it add. Consequently, the resultant points along the internal angle bisector, halfway between the forces. Options B and C contradict the symmetry, and option D would require a zero resultant, which does not occur here.

The governing concept is superposition together with geometric symmetry. The centre is at the same distance from all four corners, so each corner produces an equal repulsive force on the central positive charge. The force from each corner is cancelled by the equal and opposite force from the corner directly across the square. Both opposite pairs cancel, leaving a net force of zero. A directional answer would violate the fourfold symmetry.

Use the direction rules for electrostatic forces. The positive charge above repels the positive central charge, so its force is downward. The negative charge below attracts the central positive charge, and that force is also downward. Since both equal-distance interactions point in the same direction, their magnitudes add rather than cancel. Therefore the net force is downward. It is not zero, upward, or horizontal because no force points oppositely to the downward pair.

Coulomb’s law gives the force magnitude as F = k|qQ|/r². The upper positive charge attracts the central negative charge upward, while the lower positive charge attracts it downward. Because the two source charges have equal magnitudes and equal distances, the two forces are equal. They act in opposite directions and cancel vectorially, so the net force is zero. Thus option C is correct; an upward or downward answer ignores the cancelling partner.

Use the attraction–repulsion rule together with Coulomb’s law. The negative charge on the left attracts the positive target toward the left. The positive charge on the right repels the positive target away from itself, which is also toward the left. Since both forces point left and have equal magnitudes under the stated conditions, they add rather than cancel. Therefore the net force is leftward, making option A correct; zero would require opposite directions.