Coulomb's Law (force between multiple charges)

Use Coulomb’s force rules separately for the two source charges. The positive source above repels the positive target, pushing it downward. The negative source below attracts the positive target, also pulling it downward. Equal distances and equal magnitudes make the two force magnitudes equal, but their directions are the same, so they add rather than cancel. The net force is therefore downward, making option B correct; there is no horizontal component.

For the negative target, the positive charge above attracts it upward. The negative charge below repels it away from that source, which is also upward. Because the source charges have equal magnitudes and equal distances from the target, the two force magnitudes are equal; since both point upward, they add. The resultant is therefore upward. It is not zero because cancellation requires opposite directions, so option A is correct.

Let the side of the square be a. The adjacent-corner separation is a, while the opposite-corner separation is the diagonal a√2. Coulomb’s law gives F ∝ 1/r², so F_opposite/F_adjacent = a²/(a√2)² = a²/(2a²) = 1/2. The opposite charge therefore produces half the force of an adjacent charge. The equal charge magnitudes cancel in the ratio, making option A correct.

The centre is equidistant from all four corners, and all source-charge magnitudes are equal, so each individual force on the target has the same magnitude. Charges at opposite corners form pairs: the force from one member of a pair is exactly opposite to that from the other member. This cancellation occurs for the positive pair and independently for the negative pair, regardless of the target’s sign. Hence the vector sum is zero, so option A is correct.

Coulomb’s law gives F = kq₁q₂/r². Tripling one charge multiplies the force by 3, while making the distance six times larger divides it by 6² = 36. Thus the overall factor is 3/36 = 1/12. The new force is 108 × 1/12 = 9 N, so option B is correct. The other values do not apply both scaling factors correctly.

By Coulomb’s law, F is proportional to q₁q₂/r². Multiplying both charges by 5 multiplies their product by 25, while multiplying the distance by 3 divides the force by 9. Therefore F′/F = 25/9. Substitution gives F′ = 145.8 × 25/9 = 405 N. Hence B is correct; the other choices do not equal the required scaling result.

Coulomb’s law states F ∝ q₁q₂/r². If the charge product is multiplied by 25 but F is to stay constant, r² must also be multiplied by 25. Taking the positive physical distance ratio gives r′/r = √25 = 5. Therefore the distance must be made five times larger, so A is correct. A would incorrectly use the product itself rather than its square-root relation.

Coulomb’s law gives F = k(q₁q₂)/r². Using ratios, F′/F = 16(r/r′)². We require F′/F = 4, so 4 = 16(r/r′)², which gives (r′/r)² = 4. Since distance is positive, r′/r = 2. Therefore the distance must double, making option B correct; leaving it unchanged would make the force sixteen times larger.

The direction of a vector resultant follows the symmetry of its components. For two forces of equal magnitude, the components perpendicular to the angle bisector cancel, while the components along the bisector add. Therefore the resultant points along the internal angle bisector, halfway between the two 20 N forces. Its magnitude would be 2F cos15°, but that value is not needed for the direction. Thus A is correct.

For two equal vectors of magnitude F separated by angle θ, the resultant is R = √(F² + F² + 2F² cosθ) = 2F cos(θ/2). With θ = 120°, R = 2F cos60° = F. Since R = 24 N, each force is F = 24 N. Therefore option B is correct; 12 N and 48 N would give different resultants.

At the selected vertex, the two repulsive forces each have magnitude 40 N. The angle between their directions is 60°, because the triangle is equilateral. For equal vectors, R = √(F² + F² + 2F²cos60°) = F√3. Therefore R = 40√3 N, directed along the angle bisector away from the triangle. Hence C is correct; direct addition would incorrectly give 80 N.

The two adjacent sides of a square meet at 90°, so the two forces from the adjacent corners are perpendicular. Applying the Pythagorean theorem to equal perpendicular vectors gives R = √(15² + 15²) = 15√2 N. The resultant points along the angle bisector between the two forces. Thus C is correct; 30 N would be valid only for parallel forces in the same direction.

Use Coulomb’s proportionality F ∝ q/r² because the target charge and constant k are common. The left charge produces a rightward repulsion proportional to 12/2² = 12/4 = 3. The right charge produces a leftward repulsion proportional to 3/1² = 3. These equal forces act in opposite directions, so their vector sum is zero. Therefore A is correct.

Coulomb’s law gives force proportional to the product of charge magnitudes divided by the square of separation. Both outer charges are positive, so the left charge repels the positive target to the right, while the right charge repels it to the left. The left-side factor is 32/4² = 2, whereas the right-side factor is 6/2² = 1.5. Since 2 is larger, the resultant force is rightward. Thus option A is correct; zero would require equal opposite forces.

The middle positive charge is repelled by both outer positive charges. Initially the distances are equal, so equal outer charges would produce equal and opposite forces. Reducing the left charge to one fourth reduces its repulsive force to one fourth, while the right charge and its distance remain unchanged. Therefore the leftward repulsion from the right charge is larger than the weakened rightward repulsion from the left charge. The net force is leftward, so option A is correct; it is not zero.

A negative charge is attracted toward each positive outer charge. With equal original charges and equal distances, the two attractions would cancel. When the left positive charge becomes five times larger, the leftward attraction becomes five times its former value, while the rightward attraction remains unchanged. Consequently the leftward force is greater than the rightward force, so the resultant points left. Option A is correct; zero would apply only if the two attractions remained equal.

The positive target is attracted toward each negative charge. The left charge pulls it leftward with a relative factor 9/3² = 1, while the right charge pulls it rightward with factor 4/2² = 1. These forces have equal magnitudes and opposite directions, so their vector sum is zero. Therefore option A is correct. A directional answer would require unequal force magnitudes, and “twice right” is unsupported because the two calculated factors are equal.

The left source charge is negative like the negative target, so it repels the target to the right. Its relative force factor is 8/2² = 2. The positive charge on the right attracts the negative target toward itself, also to the right, with factor 1/1² = 1. Since both forces point rightward, their magnitudes add rather than cancel. Thus the net force is rightward, making option A correct; zero and leftward directions contradict the force directions.

Coulomb’s law states that force is proportional to 1/(εr²) when the charges are fixed. The medium alone reduces the original 200 N force to 200/4 = 50 N. Halving the separation changes the inverse-square factor by 1/(1/2)² = 4, multiplying that value by four: Fnew = 50 × 4 = 200 N. Therefore option B is correct; the medium reduction and distance increase exactly cancel.

From Coulomb’s law, F is proportional to q1q2/r². Let the new distance be xr. Then F'/F = 12/x². The required force ratio is 3, so 12/x² = 3, giving x² = 4 and x = 2 for a positive distance. Therefore the separation must be doubled, making option A correct. Halving the distance would increase the force by an additional factor of four, not produce the required factor of three.

Using F = kq1q2/r², let the distance factor be x. The new-to-old force ratio is (2/3)/x², and it must equal 1/6. Thus (2/3)/x² = 1/6, so x² = 4 and x = 2 because distance is positive. The separation must therefore double, giving option A. Tripling would reduce the force too much, while halving would increase it rather than reduce it.

Take the target charge as positive and compare force factors q/r². The positive charge on the left repels the target to the right with factor 8/2² = 8/4 = 2. The negative charge on the right attracts the positive target to the right with factor 16/4² = 16/16 = 1. Both contributions point rightward, so their resultant also points rightward. Option A is correct; cancellation is impossible because the directions are the same.

Because the target is positive, each negative source attracts it. The left source therefore pulls left with a relative factor 8/2² = 8/4 = 2. The right source pulls right with factor 32/4² = 32/16 = 2. These forces have equal magnitudes but opposite directions, so their vector sum is zero. Hence option C is correct. The signs determine attraction, while the q/r² comparison establishes the exact cancellation.

The negative target is attracted toward each positive source. The upper source pulls upward with relative factor 9/3² = 9/9 = 1. The lower source pulls downward with factor 1/1² = 1. Since the upward and downward forces are equal in magnitude and opposite in direction, the vector resultant is zero. Therefore option A is correct; neither upward nor downward force dominates.

Let the side of the square be a. An adjacent corner is at distance a, while the opposite corner is at the diagonal distance a√2. Coulomb’s law gives F ∝ 1/r². Thus F_opposite/F_adjacent = a²/(a√2)² = a²/(2a²) = 1/2. Therefore the force due to the opposite corner is half the force due to an adjacent corner. Option A is correct; the diagonal is not twice the side.