Coulomb's Law (force between multiple charges)

Coulomb’s law gives F = kq₁q₂/r². Multiplying one charge by 4 multiplies the force by 4, while multiplying the distance by 3 divides it by 3² = 9. Thus F′ = 72 × 4/9 = 32 N. Option A would result from using an incorrect distance factor, and options C and D ignore the inverse-square effect or combine the changes incorrectly.

Use Coulomb’s relation F ∝ q₁q₂/r². Halving both charges changes their product to (1/2)(1/2) = 1/4 of its original value. Halving the distance changes 1/r² by a factor of 1/(1/2)² = 4. The total factor is (1/4) × 4 = 1, so F′ = 96 N. Therefore option C is correct; the other values result from applying only one change or using the wrong distance dependence.

Coulomb’s law can be written as F ∝ (q₁q₂)/r². A sixteenfold increase in the charge product multiplies F by 16. A fourfold increase in separation multiplies r² by 4² = 16, so it divides F by 16. The combined factor is 16/16 = 1; hence the force remains unchanged. Options B, C, and D consider only one factor or combine the factors incorrectly.

From Coulomb’s law, F ∝ q/r² when the other charge is fixed. Halving one charge contributes a factor of 1/2. Increasing the separation fourfold contributes 1/4² = 1/16. Therefore F′ = 128 × (1/2) × (1/16) = 128/32 = 4 N. Option B misses one factor of two, while C and D do not apply the inverse-square distance dependence correctly.

The vector resultant of two forces P and Q separated by angle θ is R = √(P² + Q² + 2PQ cosθ). Here P = Q = 14 N and θ = 90°, so cos90° = 0. Hence R = √(14² + 14²) = √392 = 14√2 N. Direct addition gives 28 N only for parallel forces in the same direction; zero applies to equal opposite forces.

For two equal forces F making an angle θ, R = √(F² + F² + 2F²cosθ). At θ = 60°, cosθ = 1/2, so R = √(3F²) = F√3. Given R = 20√3 N, division by √3 gives F = 20 N. Option A is half the required value, C doubles it, and D mistakes the resultant for each individual force.

The selected charge is repelled by the other two positive charges. The two force vectors each have magnitude 30 N, and the angle between them equals the triangle’s interior angle, 60°. Therefore R = √(30² + 30² + 2·30·30·cos60°) = 30√3 N. Simply adding gives 60 N, while zero would require equal opposite directions, which are not present here.

The two adjacent vertices lie along the two perpendicular sides meeting at the target corner. Since all charges are positive, both forces repel the target along perpendicular directions, and each has magnitude 12 N. Thus R = √(12² + 12²) = 12√2 N. Adding magnitudes gives 24 N only when directions coincide; cancellation to zero would require opposite, not perpendicular, forces.

Apply F = k|qQ|/r² and compare the two forces on the target. The left charge contributes a relative factor 2/1² = 2 and repels the positive target to the right. The right charge contributes 8/2² = 8/4 = 2 and repels it to the left. The magnitudes are equal and directions opposite, so they cancel and the net force is zero. The left or right options ignore this exact balance.

Coulomb’s law gives the force magnitude as proportional to |Qq|/r². Since all charges are positive, the left charge repels the target toward the right, with relative magnitude 27/3² = 3. The right charge repels it toward the left, with relative magnitude 2/1² = 2. These opposite forces do not cancel; the larger rightward contribution leaves a net force toward the right. Therefore option A is correct.

The positive target is attracted toward each negative source. The left source attracts it leftward with relative strength 4/2² = 1. The right source attracts it rightward with relative strength 1/1² = 1. The two forces therefore have equal magnitudes but opposite directions, so their vector sum is zero. The result is not twice right or toward either side; option A is correct.

The left source and the target are both negative, so they repel; the left source pushes the target rightward. Its relative strength is 4/2² = 1. The right source is positive while the target is negative, so it attracts the target rightward, with strength 1/1² = 1. Since both contributions point right, they add rather than cancel. Thus option B is correct.

Coulomb’s law states that force is proportional to q₁q₂ and inversely proportional to the medium’s permittivity. Doubling both charges multiplies q₁q₂ by 2 × 2 = 4. The stated medium effect multiplies the force by 1/4. Combining the factors gives 4 × 1/4 = 1, so the force remains unchanged: 120 N. Option A ignores the charge change, while C ignores the medium reduction; option B is correct.

Coulomb’s law gives F proportional to 1/r² when the charges are fixed. The medium changes the original force by a factor of 1/5. Halving the separation changes the distance factor to 1/(1/2)² = 4. Therefore the new force is 80 × (1/5) × 4 = 64 N. Hence option B is correct; option A ignores the distance change, while C and D use an excessive increase.

Coulomb’s law is F = kq₁q₂/r², so the force depends on the charge product divided by the square of the separation. If r becomes 5r, r² becomes 25r², which would reduce the force to 1/25 unless the charge product also increases by 25. Thus q₁q₂ must become twenty-five times its original value, making option C correct.

From Coulomb’s law, F is proportional to (q₁q₂)/r². Reducing the charge product to one-fourth already reduces the force to one-fourth if the separation is unchanged. Since the required final force is exactly one-fourth, no additional distance factor is allowed: r must remain r. Changing the distance would produce a different result, so option A is correct.

Use the ratio form F′/F = (q₁′q₂′/q₁q₂)(r/r′)². The charge product contributes a factor of 8, but the desired force factor is only 2. Therefore (r/r′)² = 2/8 = 1/4, so r/r′ = 1/2 and r′ = 2r. Doubling the distance reduces the force by four, leaving a net factor 8/4 = 2. Option A is correct.

The left positive charge repels the positive target toward the right. Its force is proportional to q/r², giving a relative factor 1/1² = 1. The right negative charge attracts the target toward the right; its relative factor is 9/3² = 9/9 = 1. Both forces therefore point rightward and have equal magnitude, so their resultant is also to the right. Option B is correct.

Both negative charges attract the positive target. The left charge pulls left with relative strength 9/3² = 1, while the right charge pulls right with relative strength 1/1² = 1. These forces are equal in magnitude and opposite in direction, so they cancel by the superposition principle. The net force is zero. Options B and D incorrectly treat the left force as larger, while C reverses the dominant direction.

A negative target is attracted toward each positive source. The upper charge pulls upward with relative strength 4/2² = 1. The lower charge pulls downward with relative strength 1/1² = 1. Since the two forces have equal magnitudes and opposite vertical directions, their vector sum is zero. There is no horizontal component in the stated arrangement, so option C is the only correct answer.

For a fixed external charge and separation, Coulomb’s law gives F proportional to the target-charge magnitude. Multiplying the target magnitude by 4 therefore multiplies the force magnitude by 4. Reversing the target’s sign changes attraction into repulsion or repulsion into attraction, so the force direction reverses for every external charge. Hence option A is correct; the other choices miss one or both effects.

Initially, the equal outer positive charges exert equal repulsive forces on the middle charge: the left charge pushes it right and the right charge pushes it left. When the right charge is halved, its leftward force becomes half as large, whereas the left charge’s rightward force is unchanged. The rightward force therefore dominates, so the net force is toward the right. Option B is correct.

Like charges repel. Thus, the enlarged left negative charge repels the middle negative charge toward the right, while the unchanged right negative charge repels it toward the left. Because the distances are equal, the force ratio is determined by the charge magnitudes: the rightward force has factor 5 and the leftward force factor 1. The resultant is therefore rightward, making option A correct.

The sign of q₁q₂ determines whether the Coulomb force is attractive or repulsive, while its magnitude is proportional to |q₁q₂|. Initial attraction means the charges have opposite signs. Reversing only one sign makes them have the same sign, so the force becomes repulsive. Tripling both magnitudes changes the product by 3 × 3 = 9, with distance unchanged. Hence the new force is nine times the original magnitude and repulsive; option B is correct.

Repulsion initially means that the two charges have the same sign. Reversing the sign of one charge makes their signs opposite, so the interaction changes from repulsive to attractive. Coulomb’s-law magnitude is proportional to the product of the charge magnitudes. Increasing only one charge by a factor of four therefore makes the force magnitude four times the original, assuming the separation is unchanged. Thus option A is the unique correct answer.