Coulomb's Law (force between multiple charges)

The negative charge above repels the negative target, so the target is pushed downward, away from the upper charge. The positive charge below attracts the negative target, pulling it downward toward the lower charge. Equal distances and equal magnitudes make the two force magnitudes equal, and both vectors point downward. They therefore reinforce one another instead of cancelling. Hence the net force is downward, so option B is correct.

Coulomb’s law gives F = kq₁q₂/r². Multiplying one charge by 6 multiplies the force by 6, while multiplying distance by 9 divides it by 9² = 81. Thus F′/F = 6/81 = 2/27, so F′ = 162 × 2/27 = 12 N. Hence B is correct. The other values result from using the distance factor incorrectly or ignoring the square dependence.

Use Coulomb’s relation F = kq₁q₂/r². Tripling both charges makes q₁q₂ nine times larger. Increasing the separation six times makes r² thirty-six times larger. Therefore F′/F = 9/36 = 1/4, and F′ = 240/4 = 60 N. Option A is correct. The other choices come from treating the distance as a first-power factor or applying only one of the two changes.

Coulomb’s law states F ∝ q₁q₂/r². If the charge product becomes 64 times but F is unchanged, the square of the distance must also become 64 times: r′²/r² = 64. Taking the positive physical distance gives r′/r = √64 = 8. Thus the distance must become eight times, so A is correct. A factor of 64 would be appropriate for r², not for r itself.

From F = k(q₁q₂)/r², the change factor is F′/F = 36/(r′/r)². The required force factor is 9, so 9 = 36/(r′/r)². Hence (r′/r)² = 4 and r′/r = 2, taking the positive distance ratio. Therefore the distance must double, making A correct. Keeping distance unchanged would give 36 times the force, not 9 times.

At a right angle, the two force vectors are perpendicular, so the Pythagorean theorem gives R = √(F² + F²) = F√2 for equal magnitudes F. Substituting R = 20√2 N yields F = (20√2)/√2 = 20 N. Hence B is correct. Option D confuses the resultant with one component, while 10 N and 40 N give different resultants.

The adjacent sides of a square meet at 90°, so the two specified 20 N force vectors are perpendicular. Their resultant follows the Pythagorean theorem: R = √(20² + 20²) = √800 = 20√2 N. Therefore B is correct. Direct addition to 40 N would apply only to parallel forces in the same direction, while cancellation would require equal opposite directions.

Coulomb’s law gives force proportional to the product of charge magnitudes divided by the square of separation, F ∝ Q/r². The left charge gives a relative factor 80/4² = 80/16 = 5 and repels the positive target to the right. The right charge gives 12/2² = 12/4 = 3 and repels it to the left. Since 5 exceeds 3, the net force is rightward. Thus option A is correct; it is not zero because the opposing forces are unequal.

The governing concept is Coulomb’s law together with superposition. Initially, equal outer charges at equal distances exert equal and opposite forces on the positive middle charge. The left charge still repels the middle charge to the right with the original force F. The reduced right charge produces only F/5 and repels it leftward. Therefore the net force is F − F/5 = 4F/5 toward the right. Option A is correct; it is not zero because the symmetry has been broken.

A negative charge is attracted toward each positive outer charge. Because the distances from the midpoint are equal, Coulomb’s law shows that each force is proportional to the corresponding outer charge. The tripled left charge produces a force 3F toward the left, while the halved right charge produces F/2 toward the right. Their vector difference is 3F − F/2 = 2.5F leftward. Hence option A is correct; the forces do not cancel.

The positive target is attracted toward each negative charge. By Coulomb’s law, the left force has relative magnitude 16/4² = 16/16 = 1, directed leftward. The right force has relative magnitude 4/2² = 4/4 = 1, directed rightward. These equal forces act in opposite directions, so their vector sum is zero. Option A is correct; neither side dominates, and “twice right” is inconsistent with the equal magnitudes.

Use Coulomb’s law and inspect the direction of each interaction. The left charge is also negative, so it repels the negative target away from the left, producing a rightward force proportional to 18/3² = 2. The right positive charge attracts the negative target toward the right, producing another rightward force proportional to 1/1² = 1. Both contributions point right, so the net force is rightward. Option A is correct.

For two equal forces F separated by angle θ, the resultant is R = √(F² + F² + 2F² cos θ) = 2F cos(θ/2). For R to be zero, cos(θ/2) must be zero, which gives θ/2 = 90° and therefore θ = 180°. At this angle the forces have equal magnitudes and opposite directions, so they cancel exactly. Thus option D is correct; 90° gives a nonzero resultant of √2F.

Apply the two stated effects multiplicatively. The medium changes the force from F to F/6 because the permittivity effect reduces it. Halving the separation increases a Coulomb force by the inverse-square factor 1/(1/2)² = 4. Therefore the new force is 300 × (1/6) × 4 = 300 × 2/3 = 200 N. Option C is correct. Using only the medium factor gives 50 N, while using only the distance factor gives 1200 N.

Coulomb’s law states that force is proportional to the product of the charges and inversely proportional to the medium’s permittivity. The medium changes the original force by a factor of 1/9, while tripling one charge multiplies the force by 3. Therefore, the combined factor is (1/9) × 3 = 1/3. Hence the new force is 118/3 = 39.3 N approximately, so option B is correct. Option A uses an incorrect factor.

Coulomb’s law gives F = kq₁q₂/r². Let the original charge product be Q and distance be r. When distance becomes 11r, the denominator becomes (11r)² = 121r². To keep F unchanged, the numerator q₁q₂ must also be multiplied by 121. Thus the required charge product is 121 times its original value, making option C correct. Multiplying by only 11 ignores the square dependence on distance.

For any fixed external charge and separation, Coulomb’s force is directly proportional to the target charge: F ∝ q. Increasing the target’s magnitude by 7 multiplies the magnitude of every individual force by 7. Reversing its sign changes attraction into repulsion or repulsion into attraction, so the direction of each force reverses relative to its previous direction. Therefore option A is the only complete statement; the exact direction depends on the external charge’s sign.

For two forces of equal magnitude F separated by angle θ, the resultant is R = √(F² + F² + 2F²cosθ) = 2F cos(θ/2) over this range. At θ = 0°, the forces act in the same direction and R = 2F, the maximum. At θ = 180°, they oppose and cancel, giving R = 0, the minimum. Therefore option A is correct.

Coulomb’s law gives F = kq₁q₂/r². With both charges fixed, F is inversely proportional to r². If the new force is 121F, then 121F/F = r²/r′², so r′² = r²/121 and r′ = r/11. Thus the new distance must be one-eleventh of the original distance, making option B correct. Increasing distance would reduce, not increase, the force.

By Coulomb’s law, F = kq₁q₂/r², so with fixed charges the force varies as 1/r². Let the new distance be r′. For F′ = F/144, we require F′/F = r²/r′² = 1/144. Hence r′² = 144r² and r′ = 12r. Therefore the distance must become twelve times the original, so option A is correct. A smaller distance would increase the force instead.

Coulomb’s law states F = kq₁q₂/r². Multiplying one charge by 49 would multiply the force by 49 if the distance stayed fixed. To compensate, the factor r² must also increase by 49, so the distance must increase by √49 = 7. Therefore r′ = 7r and option A is correct. A 49-fold distance change would reduce the force far too much because distance is squared.

The governing concept is vector equilibrium: the vector sum of all forces must be zero. The resultant of the perpendicular 9 N and 12 N forces is √(9² + 12²) = √225 = 15 N. The third force must be equal in magnitude and opposite in direction to this resultant. Hence its magnitude is 15 N, so option B is correct; 21 N is an incorrect arithmetic sum.

Coulomb’s law gives F = kq₁q₂/r². The distance changes from 4 to 8, so it doubles and the force would become one-fourth if the charges stayed unchanged. Let each equal charge be multiplied by x; their product becomes x² times larger. For the force to remain unchanged, x²/4 = 1, so x = 2. Therefore option A is correct.

Using Coulomb’s law, F = kq²/r² for equal charges. If the distance becomes r/4, the distance factor alone increases the force by 16. To compensate, the product q² must become 1/16 of its original value. Therefore each equal charge must be multiplied by 1/4, because (1/4)² = 1/16. Thus option A is correct.

Coulomb’s law determines both magnitude and nature: magnitude is proportional to |q₁q₂|, while unlike signs attract and like signs repel. Initial attraction means the charges had opposite signs. Reversing one sign makes them like-signed, so the force becomes repulsive. Multiplying one magnitude by 8 multiplies the force magnitude by 8. Hence option B is correct.