Coulomb's Law (force between multiple charges)

For fixed charges, Coulomb’s law is F = k|q₁q₂|/r², so force varies inversely as the square of separation. If the new distance is r′ = 4r, then F′/F = r²/(4r)² = 1/16. Hence only one-sixteenth of the original force remains. The answer is not one-fourth because the square of the distance, rather than the distance itself, controls the change.

Coulomb’s law states that, for unchanged charges, F is proportional to 1/r². Let the original separation be r and the new separation be r′ = r/4. Then F′/F = (r/r′)² = [r/(r/4)]² = 4² = 16. Therefore the force becomes sixteen times its original value. The distance factor must be squared, which rules out four times and one-fourth.

With separation fixed, Coulomb’s law gives F = k|q₁q₂|/r², so the force is directly proportional to each charge individually. If q₁ changes to 4q₁ while q₂ and r stay unchanged, then F′ = k|4q₁q₂|/r² = 4F. Thus the force becomes four times its original magnitude. The factor is not two, and unchanged distance does not prevent the force from changing when a charge changes.

Coulomb’s law is F = k|q₁q₂|/r². Since r is unchanged, only the product of the charges matters for the ratio. Replacing q₁ by 4q₁ and q₂ by 4q₂ gives F′/F = (4q₁)(4q₂)/(q₁q₂) = 16. Therefore the force becomes sixteen times the original value. Multiplying the factors as 4 + 4 or selecting only one factor would be incorrect.

For a fixed separation, Coulomb’s law shows that F is directly proportional to the product q₁q₂. If one charge changes from q₁ to q₁/2 while the other charge and distance remain fixed, then F′ = k|(q₁/2)q₂|/r² = F/2. Consequently, the force magnitude becomes half its original value. It does not double or become four times because only one factor in the charge product was halved.

Coulomb’s law gives F = k|q₁q₂|/r². With r unchanged, replace both charges by q₁/2 and q₂/2. The new-to-old ratio is F′/F = [(q₁/2)(q₂/2)]/(q₁q₂) = 1/4. Thus the force becomes one-fourth of its original magnitude. Halving both charges does not merely halve the force; the two factors multiply, producing one-half times one-half.

Coulomb’s law states that the magnitude of force is F = k|q1q2|/r². If both charges are doubled, the product q1q2 becomes four times larger. If the distance becomes 3r, the distance-squared factor becomes 9r², reducing the force to one-ninth. Therefore the new force is 4/9 of the original force, so option A is correct; the other ratios do not apply both changes correctly.

By Coulomb’s law, F is proportional to q1q2/r². Doubling only one charge multiplies the force by 2. Replacing r with r/2 makes the inverse-square factor increase by 1/(1/2)² = 4. Multiplying these independent effects gives 2 × 4 = 8. Thus the new force is eight times the original force, making option C correct; option B ignores the charge change.

The net force is the vector sum of all applied forces. When two forces act along the same straight-line direction, their magnitudes add directly: Fnet = 3 N + 7 N = 10 N. Therefore option B is correct, and the resultant acts in the common direction. Four newtons would be the difference for opposite directions, 21 N is an incorrect multiplication, and zero would require equal opposite forces.

Forces along one line must be added algebraically with a chosen sign convention. Take rightward as positive: Fnet = +3 N − 8 N = −5 N. The negative sign means the resultant points leftward, and its magnitude is 5 N. Hence option A is correct. Adding the magnitudes gives 11 N incorrectly, while option B reverses the direction and option D ignores the unequal forces.

Because the forces act in opposite directions, their resultant magnitude is found by subtracting the smaller force from the larger one: Fnet = 9 N − 4 N = 5 N. The larger force is rightward, so the resultant also points rightward. Option A is therefore correct. Option C incorrectly adds opposite forces, option B chooses the wrong direction, and option D would require equal opposing forces.

Force is a vector quantity, so both magnitude and direction must be considered when finding the net force. If two forces have equal magnitudes but point in opposite directions, their vector sum is F + (−F) = 0. Thus option A correctly explains the cancellation. The charge need not become zero, distance is irrelevant to this vector addition, and force is not scalar, so B, C, and D are incorrect.

Each negative charge attracts the central positive charge. The left charge therefore pulls it leftward, while the right charge pulls it rightward. Since the two source charges have equal magnitudes and equal distances, Coulomb’s law gives equal force magnitudes. These opposite vectors cancel, so the net force is zero and option C is correct. A or B would result only if one force were stronger, and D has no supporting vertical component.

Both negative charges attract the central positive charge, but Coulomb’s law gives F ∝ 1/r². The closer left charge therefore produces a stronger attraction than the farther right charge, assuming their magnitudes are equal as implied. The leftward force exceeds the rightward force, leaving a net force toward the left; option A is correct. A zero result would require equal distances, while upward has no component in this arrangement.

The central negative charge is attracted toward both positive charges. Because the right positive charge is closer and the charges have equal magnitudes, Coulomb’s inverse-square law makes the rightward attraction stronger: F ∝ 1/r². The two forces do not cancel; the larger rightward force determines the resultant direction. Therefore option B is correct. Zero would require equal distances, and leftward would incorrectly favour the farther charge.

Like charges repel, so the closer left positive charge pushes the central positive charge away from the left, namely toward the right. The right positive charge pushes it toward the left, but because it is farther away, its force is weaker according to F ∝ 1/r². The stronger rightward force remains after vector addition, so option A is correct. Equal distances would produce cancellation; upward is impossible in this collinear setup.

The governing concept is Coulomb’s inverse-square law together with vector addition of forces. Like charges repel, so the left charge pushes the middle charge to the right, while the closer right charge pushes it to the left. Because force varies as 1/r², the smaller distance makes the leftward force larger. Therefore the resultant force is toward the left, not zero or downward.

Force is a vector quantity, so both magnitude and direction must be considered when combining forces. For perpendicular forces F₁ and F₂, the resultant magnitude is R = √(F₁² + F₂²), and its direction is found from tan θ = F₂/F₁. Thus vector addition is required. Simple subtraction or selecting one force ignores the directional information.

Coulomb’s law for two point charges is F = k|q₁q₂|/r², where r is the separation between their centres. Therefore, with charges unchanged, force is inversely proportional to the square of distance. If distance becomes twice as large, the force becomes one-fourth; if distance becomes three times, it becomes one-ninth. Hence option C is correct.

Coulomb’s law is F = k|q₁q₂|/r². When the separation and medium remain fixed, the magnitude of force is directly proportional to the product |q₁q₂| of both charges. Doubling either one charge doubles the force, while doubling both charges makes it four times larger. It does not depend on only one charge, their sum, or neither charge, so option A is correct.

The superposition principle states that the total electrostatic force on a charge equals the vector sum of the individual forces produced by all other charges. First calculate each pairwise force using Coulomb’s law, including its direction, and then add the vectors. The nearest charge is not automatically the only contributor, and distances cannot be ignored. Therefore option A correctly describes the principle.

The sign of q₁q₂ determines the nature of the electrostatic interaction. If q₁q₂ is negative, the charges have opposite signs and Coulomb’s force is attractive; one charge is positive and the other negative. Equal signs produce repulsion, while a zero charge would produce no electrostatic force. Attraction alone does not show which charge is positive, so option B is the only valid conclusion.

According to Coulomb’s law, two nonzero charges repel when their product q₁q₂ is positive. A positive product means the charges have the same sign: both may be positive or both may be negative. Opposite signs cause attraction, and a zero charge cannot produce repulsion. The distance affects the magnitude through 1/r² but does not change the sign-based nature of the force. Thus option B is correct.

For two point charges, the magnitude is F = k|q₁q₂|/r². When the medium and distance r are the same for both pairs, k/r² is a common factor. The pair with the larger absolute product |q₁q₂| therefore experiences the greater force magnitude. The sign of the product determines attraction or repulsion, but the comparison of magnitudes uses its absolute value. Hence option A is correct.

Coulomb’s law gives F = k|q₁q₂|/r². Since the charge product is equal for both pairs, the only changing factor is r². A smaller separation produces a larger force because the denominator is smaller; for example, halving the distance makes the force four times larger. The signs determine attraction or repulsion, not this magnitude comparison. Therefore option B is correct.