Coulomb's Law (force between multiple charges)

Coulomb’s law gives F = kq₁q₂/r². Doubling both charges changes the charge product by 2 × 2 = 4, while making the distance four times changes the denominator by 4² = 16. Hence the new force is F′ = F(4/16) = 40/4 = 10 N. Therefore option C is correct. The larger charge product increases the force, but the squared distance effect is stronger; the other values do not apply the combined ratio correctly.

Use Coulomb’s proportionality F ∝ q₁q₂/r². The charge-product factor is 3 × 1/2 = 3/2. Doubling the distance contributes a factor 1/2² = 1/4. Therefore F′/F = (3/2)(1/4) = 3/8, so F′ = 88 × 3/8 = 33 N. Option A is correct. The other choices result from missing either the halved charge or the squared distance dependence.

Coulomb’s law states F = k(q₁q₂)/r². Increasing the charge product ninefold multiplies the force by 9. Increasing the distance threefold multiplies r² by 3² = 9, so it divides the force by 9. The combined factor is 9/9 = 1; therefore the force remains unchanged and option C is correct. Options A and B ignore the inverse-square distance effect, while D reverses the net result.

From Coulomb’s law, F ∝ q₁q₂/r². Multiplying each charge by 5 makes the product 25 times larger. Multiplying the separation by 5 makes r² 25 times larger as well. Thus F′/F = 25/25 = 1, and F′ = 125 N. Option B is correct. A would incorrectly retain only the distance effect, while C and D fail to divide by the squared change in separation.

The governing concept is vector addition. For two forces of magnitudes F₁ and F₂ separated by angle θ, R = √(F₁² + F₂² + 2F₁F₂cosθ). Substituting F₁ = F₂ = 10 N and cos60° = 1/2 gives R = √(100 + 100 + 100) = √300 = 10√3 N. Thus A is correct; direct addition gives 20 N only for the same direction, and zero requires opposite forces.

Apply the vector-resultant relation R = √(F² + F² + 2F²cos120°). Since cos120° = −1/2, R² = 2F² − F² = F², so R = F. With F = 18 N, the resultant is 18 N. Therefore option C is correct. Zero would require forces at 180°, and 36 N would occur when they point in the same direction, not at 120°.

At the selected vertex, the two repulsive Coulomb forces each have magnitude 20 N and the angle between them is 60°, the interior angle of the equilateral triangle. Therefore R = √(20² + 20² + 2·20·20·cos60°) = √1200 = 20√3 N. Option B is correct. The forces do not add to 40 N because they are not parallel, and they do not cancel because their directions are not opposite.

Coulomb’s force acts along the straight line joining the source charge and the target charge. From the fourth corner of a square, the two adjacent corners lie along the two perpendicular sides meeting at that corner. Both source charges are positive, so the forces on the positive target are repulsive and point away along those perpendicular sides. Hence the angle between the forces is 90°, making option C correct.

Coulomb’s law gives F = k|qQ|/r². The left charge produces a rightward force proportional to 1/1² = 1. The right charge produces a leftward force proportional to 4/2² = 4/4 = 1. The forces have equal magnitudes but opposite directions, so they cancel and the net force is zero. Thus option C is correct; options A and B ignore one opposing contribution, while D is impossible for collinear charges.

By Coulomb’s law, each force is proportional to source charge divided by the square of its distance. The left charge pushes the positive target to the right with relative magnitude 9/2² = 9/4. The right charge pushes it left with relative magnitude 1/1² = 1. Since 9/4 is larger, the rightward force remains after subtraction, so option B is correct. The forces are not equal, so zero is not possible.

Both source charges are positive and the target is negative, so both interactions are attractive. The left source pulls the target left with relative strength 4/2² = 1. The right source pulls it right with relative strength 1/1² = 1. Since the magnitudes are equal and directions are opposite, their vector sum is zero. Therefore option C is correct; the negative sign changes attraction versus repulsion, but not the equality calculation.

For two vectors with angle θ, the resultant satisfies R² = F₁² + F₂² + 2F₁F₂cosθ. With F₁ = F₂ = 10 N and θ = 60°, R² = 100 + 100 + 200(1/2) = 300, so R = 10√3 ≈ 17.32 N. This is greater than 10 N but less than 20 N. Thus option A is correct; equal forces do not cancel because their angle is not 180°.

Coulomb’s law states that force is proportional to the product of the two charge magnitudes and inversely proportional to the square of distance. Making both charges three times multiplies the force by 3 × 3 = 9. The medium then reduces this result to one-fourth, so the net factor is 9/4. Therefore, the new force is 9 × 9/4 = 20.25 N. Option B ignores the medium, while the other values use an incorrect factor.

Coulomb’s law makes the force directly proportional to either charge when the other charge and distance remain unchanged. The medium first changes the air force from 50 N to 50/5 = 10 N. Doubling one charge then doubles this value: 2 × 10 = 20 N. Thus option B is correct. Option A stops after applying only the medium factor, whereas options C and D fail to include the one-fifth reduction correctly.

Coulomb’s law is F = kq₁q₂/r². If the distance changes from r to 2r, the denominator becomes (2r)² = 4r², so the force would become one-fourth if the charge product stayed unchanged. To preserve the original force, q₁q₂ must therefore be multiplied by four. Option B is correct; doubling the product would still leave the force at half its original value.

For two point charges, Coulomb’s law gives F = kq₁q₂/r². Replacing r by 3r changes the distance factor to (3r)² = 9r², which would reduce the force to one-ninth if the charge product were unchanged. Hence q₁q₂ must also increase by a factor of nine to cancel that change and keep F constant. Option C is correct; a threefold increase compensates only the distance, not its square.

Using F = kQ/r², let the original force be F = kQ/r². After doubling the charge product and changing distance to nr, the new force is F′ = 2kQ/(n²r²) = 2F/n². The required value is F′ = F/2, so 2/n² = 1/2, giving n² = 4 and n = 2. Therefore the distance must be doubled, making option A correct.

Apply Coulomb’s law to the positive target. The positive charge on the left repels it toward the right, with relative force proportional to 1/1² = 1. The negative charge on the right attracts it toward the right; its relative magnitude is 9/3² = 9/9 = 1. Both forces therefore point rightward and have equal magnitude, so their sum is still rightward, not zero. Option B is correct.

Step 1: The left negative charge attracts the target leftward with factor four by four which is one. Step 2: The right positive charge repels the positive target leftward with factor one. Step 3: Both forces are leftward so net force is twice one force toward left.

Both external charges are negative and the target is positive, so each charge attracts the target. The left charge pulls leftward with relative magnitude 4/2² = 4/4 = 1. The right charge pulls rightward with relative magnitude 1/1² = 1. These forces are equal in magnitude and opposite in direction, so vector addition gives zero net force. Option C is correct; considering only charge size or only distance would give a wrong direction.

For each external charge, Coulomb’s force magnitude is |F| = k|q_target||q_external|/r². Changing only the target from +q to −q leaves |q_target| and every distance unchanged, so every individual force magnitude remains the same. The sign reverses the attraction or repulsion direction, and consequently the vector sum reverses direction as well, while its magnitude is unchanged. Thus option A is correct; the other statements do not follow from Coulomb’s law.

Initially, equal positive charges at equal distances repel the middle charge with equal and opposite forces, producing zero resultant. After the left outer charge is doubled, its repulsive force on the middle charge also doubles because Coulomb force is proportional to the source charge. The left charge therefore pushes the middle charge rightward more strongly than the right charge pushes it leftward. The net force is rightward, so option B is correct.

Like charges repel. Thus the left negative charge repels the middle negative charge toward the right, while the right negative charge repels it toward the left. With equal original charges and equal spacing, these forces initially cancel. Making the right outer charge four times makes its leftward force four times larger, while the rightward force from the left charge is unchanged. The resultant is therefore leftward, so option A is correct.

Let the side of the square be a. An adjacent corner is at distance a, whereas the diagonally opposite corner is at distance a√2. Coulomb’s law gives F ∝ 1/r². Therefore Fdiagonal/Fadjacent = a²/(a√2)² = a²/(2a²) = 1/2. The equal charges and unchanged medium cancel in the comparison. Thus the diagonal charge produces half the force of an adjacent charge.

The governing principle is superposition: the net force on a chosen target charge equals the vector sum of the Coulomb forces exerted on it by all other charges. Calculate each force using F = k|qQ|/r², determine attraction or repulsion and its direction, resolve components if needed, and then add vectors. Direct magnitude addition ignores direction, the largest charge need not dominate, and forces between external charges do not act on the target.