Coulomb's Law (force between multiple charges)

Force is a vector quantity, so its magnitude and direction must both be considered. If forces act along different directions, their vector components or a vector diagram must be used; simply adding magnitudes can give a wrong resultant. Forces in the same direction may be added algebraically, so option A is correct. The number of charges or knowledge of distance does not by itself decide whether scalar addition is valid.

The superposition principle states that each external charge produces its own force on the target, independently of the others. The net force is therefore F_net = F₁ + F₂ + F₃, where the addition is vector addition. Magnitudes alone cannot be added unless all forces have the same direction. Ignoring smaller forces or selecting only the largest one is unjustified. Thus option A is correct.

Choose a positive direction and use vector addition. Since the first two forces point in the same direction, their magnitudes can first be combined: F_same = F₁ + F₂. The opposite force is then subtracted algebraically, giving F_net = F₁ + F₂ − F₃ if the first direction is positive. This is efficient and preserves direction information. Therefore option A is correct; the other choices discard valid forces or directions.

In a one-dimensional charge arrangement, every force must be assigned a left or right direction before algebraic addition. Like charges repel and unlike charges attract, so the direction depends on the sign of both interacting charges and their positions. A frequent error is to add all magnitudes as positive without deciding their directions. Counting charges, checking distance, and writing units are useful steps, not the central mistake. Option A is correct.

Coulomb’s law is F = k|q₁q₂|/r². With r fixed, the force is directly proportional to the product q₁q₂. If q₁ is changed to 2q₁ while q₂ remains unchanged, the new force is F′ = k|(2q₁)q₂|/r² = 2F. Thus doubling either one charge doubles the force. Doubling distance would reduce force to one-fourth, halving both charges would reduce it, and making one zero would eliminate it. Option A is correct.

The force direction between two charges is determined by attraction or repulsion. Reversing the sign of the target charge changes the sign of the product qQ, while the positions and distance remain unchanged. A previously attractive interaction becomes repulsive, or a previously repulsive interaction becomes attractive, so the force vector reverses for each relevant external charge. Its magnitude is unchanged if only the sign changes. Hence option A is correct; the force need not become zero and distances do not change.

Coulomb’s law states that, for unchanged charges, force varies inversely as the square of separation: F ∝ 1/r². If the distance becomes 2r, the new force is F′ = F/(2²) = 36/4 = 9 N. Therefore option A is correct. Eighteen would represent halving the force, while the larger values reverse the distance effect.

Coulomb’s law gives F = kq₁q₂/r², so the force is directly proportional to the product q₁q₂ when distance is fixed. Doubling both charges changes the product to (2q₁)(2q₂) = 4q₁q₂. Hence the new force is 4 × 10 = 40 N, so option B is correct. Option A doubles only once and misses the second charge factor.

Using F = kq₁q₂/r², halving one charge contributes a factor of 1/2. Halving the distance changes r to r/2, so the distance factor is 1/(r/2)² = 4/r², a fourfold increase. Combining both independent changes gives F′/F = (1/2) × 4 = 2. Thus option A is correct.

The forces act along the same line in opposite directions, so choose left as positive or simply subtract the smaller magnitude from the larger one. The net magnitude is 12 − 5 = 7 N. Since the larger force points left, the resultant also points left. Therefore option C is correct; adding them would incorrectly ignore their opposite directions.

Use attraction and repulsion together with the line arrangement. The positive charge on the left repels the positive target away from it, so its force is to the right. The negative charge on the right attracts the positive target toward it, also to the right. The forces therefore reinforce one another, giving a rightward net force; option B is correct.

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 the right-hand side, which is also toward the left. Thus both Coulomb forces have the same leftward direction and their magnitudes add, so option A is correct. Equal magnitudes do not cause cancellation when directions agree.

Each outer positive charge repels the positive middle charge. Because the outer charges are equal and at equal distances, Coulomb’s law gives equal force magnitudes, F = kq²/r², on the middle charge. One force points left and the other points right, so their vector sum is F − F = 0. Therefore option C is correct by symmetry.

Like charges repel, so the negative charge on the left pushes the middle negative charge to the right, while the negative charge on the right pushes it to the left. Equal magnitudes and equal distances make these forces equal according to Coulomb’s law. They are opposite vectors, so the resultant is zero. Hence option A is correct, not a one-sided force.

Coulomb’s law gives F = kq₁q₂/r². Tripling both charges changes their product by 3 × 3 = 9, so this alone would make the force nine times larger. Tripling the distance changes r² by 9, reducing the force to one-ninth. Thus F' = 8 × 9/9 = 8 N, so option A is correct; the other choices retain only one of the two effects.

Use F = kq₁q₂/r². Increasing each charge threefold multiplies q₁q₂ by 9. Increasing the separation threefold multiplies r² by 9 in the denominator. Consequently, F'/F = (3 × 3)/(3²) = 9/9 = 1. The magnitude remains unchanged, so option B is correct. Options A and D consider only part of the scaling, while C reverses the net result.

The resultant is the vector sum of the two forces. For equal perpendicular vectors, the horizontal and vertical components of the resultant are equal, so its direction makes equal angles with both original forces. It therefore lies along the internal angle bisector, at 45° to each force. Option A is correct; it cannot lie along only one force because both components are nonzero and equal.

By Coulomb’s law, like charges repel and unlike charges attract. The positive charge on the left repels the positive target charge away from itself, so that force is directed rightward. The negative charge on the right attracts the positive target charge toward itself, also producing a rightward force. Thus both individual forces act to the right; their magnitudes may differ, but their directions are the same. Therefore option B is correct.

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 the right-hand side, which is also toward the left. Since both forces point leftward, they add rather than cancel. Their different charge magnitudes affect the total magnitude, but not the direction. Hence the net force is toward the left, so option A is correct.

Coulomb’s law gives the magnitude as F = k|qQ|/r², so the closer left charge produces the larger magnitude because its distance is smaller. However, the direction depends on whether the target and the external charges attract or repel. If the target has the opposite sign, the stronger force is leftward; if it has the same sign, the stronger repulsive force is rightward. Therefore option A is the only generally valid answer.

Coulomb’s law states F ∝ q₁q₂/r². Halving one charge multiplies the force by 1/2. Doubling the distance makes r² four times larger, so it multiplies the force by 1/4. The combined factor is (1/2)(1/4) = 1/8. Therefore the new force is 24/8 = 3 N. Thus option A is correct; the other values do not include both changes correctly.

Coulomb’s law is F = kq₁q₂/r². When both charges are halved, their product becomes (1/2)(1/2) = 1/4 of its original value. When the distance is halved, r² becomes 1/4 as large, so the inverse-square factor increases the force by four. The two factors cancel: (1/4) × 4 = 1. Hence the force remains unchanged, making option A correct.

The magnitude of Coulomb force depends on the magnitudes of the charges and their separation: F = k|q₁q₂|/r². Changing only the signs leaves this magnitude unchanged. However, the signs determine whether the interaction is attractive or repulsive; changing the sign relationship can reverse the force direction and change its nature. Units and charge magnitudes do not change merely because signs change. Therefore option A is correct.

For collinear forces, choose rightward as positive and add the signed forces. The two rightward forces give 5 + 3 = 8 N. The leftward force is negative, so the net force is 8 − 6 = 2 N. The positive result means the direction is rightward. Therefore the resultant force is 2 N to the right, making option A correct. The 14 N choice incorrectly adds opposing forces without signs.

Take leftward as the positive direction. The two leftward forces add to 4 + 5 = 9 N. The 7 N rightward force acts oppositely, so it must be subtracted: 9 − 7 = 2 N. Because the larger total is leftward, the resultant is 2 N toward the left. Hence option A is correct; 16 N would incorrectly add forces acting in opposite directions.