Coulomb's Law (force between multiple charges)

Coulomb's law states F = k|q1q2|/r², so force varies inversely as the square of separation. If the distance changes from r to 2r, the new force is F' = k|q1q2|/(2r)² = F/4. Thus one-fourth of the original force remains. It is not half because the inverse-square dependence requires squaring the factor of two; options B and D increase the force incorrectly.

The sign of the product q1q2 determines the nature of the electrostatic interaction: like charges repel and unlike charges attract. Both charges here are negative, so they have the same sign and exert equal, opposite-directed repulsive forces on one another. The force is not zero merely because the charges are equal, and gravity is not the only interaction present. Hence option B is correct.

For two point charges, Coulomb's law describes the magnitude, while the signs determine the direction of interaction. A positive and a negative charge have opposite signs, so their product is negative and the force is directed toward the other charge for each particle. This is an attractive force. It is not zero, repulsive, or described as parallel; those choices confuse force type with direction or omit the interaction.

Force is a vector quantity because it has both magnitude and direction. According to the superposition principle, the net force on a selected charge is the vector sum of the individual forces from all other charges: Fnet = F1 + F2 + F3 + …. Components may be added algebraically after choosing axes, but magnitudes alone are insufficient unless all forces are collinear and similarly directed. Therefore option B is correct.

The net force is found by vector addition, not by adding magnitudes without considering direction. Let the two equal forces be +F and −F along the same line. Their resultant is Fnet = +F − F = 0. Thus the charge has zero net force, although two individual forces are still acting on it. Options A and D would apply to forces in the same direction, while C has no valid basis.

When two forces are collinear and point in the same direction, their vectors have the same sign. Therefore their magnitudes add: Fnet = F1 + F2, and the resultant points in that common direction. Subtraction is required only when collinear forces oppose one another. Using only the smaller force or declaring the result zero discards part of the actual interaction. Hence option B is correct.

Coulomb's law is F = k|q1q2|/r². Keeping r constant while changing q1 to 3q1 and q2 to 3q2 changes the charge product to (3q1)(3q2) = 9q1q2. Consequently, the new force is F' = 9F. The direction and attractive or repulsive nature remain determined by the charge signs, which have not changed. Therefore option C is correct; three times ignores the change in both charges.

Coulomb’s law states that the magnitude of force is F = k|q₁q₂|/r², so force is inversely proportional to the square of separation. If the new distance is 3r, then F′ = F/3² = F/9. Thus the force becomes one-ninth of its original value. It does not become one-third because the square of distance, rather than distance alone, appears in the denominator; therefore option D is correct.

Coulomb’s law is F = k|q₁q₂|/r². With the charges fixed, changing the distance controls the force through the inverse square of r. If the new distance is r/2, then F′ = k|q₁q₂|/(r/2)² = 4F. Therefore the force becomes four times its original value. Half and double ignore the square dependence, while one-fourth would describe the opposite change, namely increasing the distance twofold.

Each outer positive charge repels the positive middle charge. Because the two outer charges have equal magnitude and are at equal distances, Coulomb’s law gives two forces of equal magnitude, q-side = kQq/r². One force points left and the other points right. Their vector sum is therefore zero. The result follows from symmetry; it is not an upward force because all three charges lie on one straight line. Hence option C is correct.

The two charges are both positive, so they repel. The left-hand charge pushes the middle positive charge away from its own position. Since the source charge lies to the left, “away from it” means toward the right. Coulomb’s law determines the magnitude, while the sign combination determines repulsion and therefore direction. The force is not zero because only one interacting charge is being considered; thus option B is correct.

A positive and a negative charge have opposite signs, so they attract each other. The middle positive charge is pulled toward the negative charge located on its right. Therefore the force on the middle charge points rightward. The force is not leftward, because that would represent repulsion, and it is not zero because unlike charges exert a nonzero electrostatic force when separated. Hence option B is the correct answer.

The charges have opposite signs, so the governing electrostatic interaction is attraction. The positive charge on the left pulls the negative middle charge toward the left. Thus the force direction is leftward. It is not rightward because attraction is toward the source charge, not away from it; it is not zero because the charges are separated and interact. Therefore option A correctly gives the force direction.

Both interacting charges are negative, so they have like signs and repel one another. The right-hand negative charge pushes the middle negative charge away from itself. Since the source charge is on the right, the direction away from it is toward the left. The force is therefore not toward the right, which would imply attraction, and it is not zero because the charges exert a finite electrostatic force. Hence option A is correct.

For two point charges, Coulomb’s law describes a central force: its line of action passes through both charges. Therefore the force acts along the straight line joining them. Like charges experience repulsion along this line, while unlike charges experience attraction along the same line. It is not always upward, circular, or perpendicular; those directions are not implied by the electrostatic interaction. Thus option A is correct.

Coulomb force is a type of mechanical force, and the SI unit of every force is the newton (N). This also follows dimensionally from F = ma, where mass is measured in kilograms and acceleration in metres per second squared, giving kg·m/s² = N. Coulomb is the unit of electric charge, metre is the unit of length, and joule is the unit of energy. Therefore option B is correct.

The governing relation is Coulomb’s law, F = k|q₁q₂|/r². If either q₁ or q₂ is zero, their product is zero, so F = 0 provided the separation is finite. Thus no electric interaction force acts between the two charges in this situation. Double, half, or infinite values would require a nonzero change in charge or distance, not a zero charge.

Newton’s third law and Coulomb’s law give the result. If charges q₁ and q₂ are separated by distance r, each experiences a force of magnitude F = k|q₁q₂|/r². The forces act on different charges, have equal magnitudes, and point in opposite directions. Therefore the action and reaction magnitudes are equal; they do not become unequal merely because the charges differ.

Coulomb’s law gives the magnitude as F = k|q₁q₂|/r². Reversing the sign of either or both charges does not change |q₁|, |q₂|, or r, so the numerical magnitude remains unchanged. The signs can change whether the interaction is attractive or repulsive when only one sign is reversed, but none of the listed changes occurs in the magnitude. Hence A is correct.

Use F = kq₁q₂/r². Doubling both charges changes the charge product to (2q₁)(2q₂) = 4q₁q₂, multiplying the force by four. Doubling the separation changes r² to (2r)² = 4r², reducing the force by a factor of four. The factors cancel: F′/F = 4/4 = 1. Therefore the force remains unchanged.

For Coulomb force, F is proportional to q/r² when the other charge is fixed. Multiplying one charge by 3 multiplies the force by 3, while multiplying distance by 3 divides it by 3² = 9. Thus F′/F = 3/9 = 1/3. The new force is one-third of the original, so option B is correct; the distance effect is not merely a division by three.

Coulomb’s law is F = kq₁q₂/r². Doubling both charges makes q₁q₂ four times larger. Replacing r by r/2 makes the denominator (r/2)² = r²/4, which makes the force four times larger again. Therefore F′/F = 4 × 4 = 16. The force becomes sixteen times the original value, not merely four or eight times.

For collinear forces, choose right as positive or compare the opposing magnitudes directly. The leftward force is 6 units and the rightward force is 2 units, so the net magnitude is 6 − 2 = 4 units toward the side of the larger force. Therefore the net force points left. It is not zero because the opposing forces are unequal, and no upward component is given.

Net force is the vector sum of all applied forces. Taking right as positive gives Fnet = +5 + (−5) = 0 force units. The equal magnitudes cancel because the directions are opposite, so the charge has no resultant force from these two forces. Ten units would result only if the forces acted in the same direction, while five units would require one force to be absent.

For each outer charge, Coulomb’s law gives F = k|qouter qmiddle|/r². Since the two outer charges have equal magnitude and the same middle charge is involved, the only differing factor is distance. The closer charge has the smaller r, and the inverse-square relation makes its force larger. Thus A is correct; equal charge magnitudes do not overcome a difference in separation.