Coulomb's Law (force between multiple charges)

According to Coulomb's inverse-square law, F = k|q₁q₂|/r². When the distance changes from r to 3r, the denominator becomes (3r)² = 9r², so the force becomes one-ninth of its original value. Therefore F' = 45/9 = 5 N, making option A correct. Option B divides only by three, whereas C and D incorrectly predict an increase when the separation becomes larger.

Coulomb's law states that force varies as 1/r² when the charges are fixed. If the new distance is r' = r/3, then (r/r')² = (r/(r/3))² = 3² = 9. The force therefore becomes nine times its original value: F' = 9 × 7 = 63 N. Option B is correct. Option A uses a first-power relation, while C and D predict a decrease instead of the required increase.

Start with F = k|q₁q₂|/r². Halving both charges changes their product to (q₁/2)(q₂/2) = q₁q₂/4, which tends to make the force one-fourth. Halving the distance changes r² to r²/4, which multiplies the force by four. These factors cancel: (1/4) × 4 = 1. Therefore the force remains unchanged, so option A is correct; the other options consider only one effect or combine them incorrectly.

Coulomb’s law states that F = kq₁q₂/r². Tripling both charges multiplies the numerator by 3 × 3 = 9. Doubling the distance changes r² to (2r)² = 4r², so the force is divided by 4. Therefore, F′/F = 9/4, and the correct result is nine-fourths times. The other ratios do not apply both charge and distance changes correctly.

Coulomb’s law gives F = kq₁q₂/r². Increasing one charge fourfold multiplies the force by 4. Doubling the separation makes the denominator four times larger, so the force is multiplied by 1/4. Combining the changes gives F′/F = 4 × 1/4 = 1. Thus the force remains unchanged. Option B ignores the squared distance dependence, while C ignores the distance change.

A resultant is determined by adding force vectors component by component. For two equal forces at 90°, the horizontal and vertical components of the resultant are equal. A vector with equal perpendicular components makes equal angles with both original forces, so it lies along the internal angle bisector, at 45° to each force. It cannot lie along only one force because the other equal force contributes a nonzero component.

Use attraction and repulsion together with Coulomb’s law. 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 charge, also toward the left. Thus both forces point left; their magnitudes add, and the fact that the right charge is twice as large only makes its leftward contribution stronger. The net force is therefore leftward, not zero or rightward.

Coulomb force acts along the line joining the charges. The left positive charge repels the positive target, so its force is toward the right. The right negative charge attracts the target, also toward the right. Since both contributions point in the same direction, their magnitudes add, regardless of which charge is larger. Hence the resultant force is rightward, making option B correct; zero would require opposing forces.

By Coulomb’s law, like charges repel, so the left positive charge pushes the middle charge to the right, while the right positive charge pushes it to the left. For equal source charges, force varies as 1/r². Because the left charge is closer, its repulsive force is larger than the opposing force from the right. The resultant is therefore rightward, so option A is correct.

Opposite charges attract, so each positive source charge pulls the central negative charge toward itself. The left charge produces a leftward force and the closer right charge produces a rightward force. Since Coulomb force varies as 1/r² and the source magnitudes are equal, the closer right charge exerts the greater force. The vector difference is rightward, so option B is correct.

Coulomb’s law gives F = k|qQ|/r². Because both external charges and the target charge are the same in magnitude, only the distance factor differs. If r₁:r₂ = 1:3, then F₁:F₂ = (1/r₁²):(1/r₂²) = r₂²:r₁² = 3²:1² = 9:1. Thus the nearer charge exerts nine times the force, so option C is correct.

For a fixed target charge and equal separation, Coulomb’s law reduces to F ∝ |q| because F = k|qQ|/r². The distances contribute the same denominator and the target charge is common to both forces. Therefore an external charge whose magnitude is three times larger produces a force three times larger. The force ratio is 3:1, so option A is correct; squaring the charge ratio would be an error.

Coulomb’s law states F ∝ |q|/r² when the target charge is common. For the first charge relative to the second, the charge factor is 9, while the distance factor is 1/3² = 1/9. Hence F₁/F₂ = 9 × (1/9) = 1, giving F₁:F₂ = 1:1. The charge increase exactly compensates for the larger distance, so option C is correct.

Electrostatic attraction indicates that the two nonzero charges initially have opposite signs. Reversing the sign of only one charge changes the sign relationship: opposite signs become like signs. Like charges repel according to Coulomb’s law, while the force magnitude remains based on the charge magnitudes and separation. Therefore the force changes from attractive to repulsive, making option B correct.

Repulsion means the two charges initially have the same sign: both positive or both negative. Reversing both signs preserves this relationship, changing (+,+) to (−,−) or (−,−) to (+,+). The product q₁q₂ remains positive, so Coulomb’s force remains repulsive, and its magnitude is unchanged if the charge magnitudes and distance are unchanged. Thus option A is correct.

The magnitude of Coulomb force is F = k|q₁q₂|/r², so it depends on charge magnitudes and separation, not on their signs. The signs determine whether the interaction is attractive or repulsive and therefore determine the force direction along the line joining the charges. Changing one sign can reverse that nature and direction while leaving the magnitude unchanged. Hence option A is correct.

For every external charge, Coulomb’s law gives Fᵢ = k|Qqᵢ|/rᵢ², so each force on the target is directly proportional to the target-charge magnitude |Q|. Doubling Q doubles every individual force vector. Since all directions and source charges remain unchanged, the vector sum also doubles: F_net' = 2F_net. Therefore option A is correct; it would not become four times because only one charge is doubled.

The governing idea is Coulomb’s law, F = k|qQ|/r², together with the fact that force direction depends on attraction or repulsion. Changing only the target charge’s sign leaves |q| unchanged, so the force magnitude for each source charge remains unchanged. However, attraction becomes repulsion, or repulsion becomes attraction; therefore each force vector reverses direction. Hence option A is correct; the other options incorrectly change the magnitude.

The governing principle is superposition applied to a specified body: the net force on a target is the vector sum of all forces exerted on that target. A force between two external charges acts on those two external charges, not directly on the target. It may affect their motion, but it is not a term in the target’s force sum. Thus A is correct; the forces are not necessarily zero, gravitational, or dimensionally different.

The governing concept is vector addition. Two forces cancel exactly only when their magnitudes are equal and their directions are opposite, meaning the angle between them is 180°. If equal forces make any other angle, their resultant has a nonzero magnitude; for angle θ it is R = √(F² + F² + 2F²cosθ). Therefore zero net force cannot be concluded, so B is correct. Charge sign does not alter this vector condition.

The governing idea is that the force on a charge due to another point charge acts along the line joining the two charges. At the selected vertex, the two force vectors therefore lie along the two sides meeting at that vertex. Every interior angle of an equilateral triangle is 60°, so the angle between the two force directions is 60°. Both forces are repulsive, but they point away along those same side lines. Hence B is correct.

Use the signs of the interacting charges to determine each direction. The top positive charge repels the positive central charge, so its force on the target is downward. The bottom negative charge attracts the positive central charge, so it also pulls the target downward. Equal distances and equal charge magnitudes make the two force magnitudes equal, but their common direction is downward. The resultant is therefore downward, so B is correct; cancellation would require opposite directions.

The governing relation is Coulomb’s law, F = k|qQ|/r². For the same target charge, compare each source’s charge magnitude divided by the square of its distance: F₁/F₂ = (|Q₁|/r₁²)/( |Q₂|/r₂²). A nearby small charge can therefore exert a larger force than a distant large charge if its distance advantage is sufficient. Sign determines attraction or repulsion, not magnitude here. Hence A is correct.

Apply Coulomb’s law while keeping the target charge and the original reference conditions fixed: F = kqQ/r². If Q becomes 4Q, the numerator makes the force four times larger. If r becomes 2r, the inverse-square factor makes it 1/(2²) = 1/4 of the original value. Combining both changes gives F′ = k(4Q)q/(2r)² = 4F/4 = F. Therefore A is correct; the effects exactly balance.

Coulomb’s law states that the magnitude of electrostatic force is proportional to the product of the charges and inversely proportional to the square of separation: F ∝ q/r². If the distance becomes 3r, the force due to the same charge becomes F/9. Therefore the charge must be increased by a factor of 9 to restore the original force. Three times would not compensate for the squared-distance effect, while one third would reduce the force further.