Coulomb's Law (force between multiple charges)

The force on a charge depends on whether the other charge attracts or repels it. The positive charge on the left attracts the negative target toward the left. The negative charge on the right repels the negative target away from the right-hand source, also toward the left. Thus the two equal-distance contributions have the same direction and combine to give a leftward net force. Option A is correct; cancellation would require opposite force directions.

Both external charges are positive, so each repels the positive middle charge. The left source pushes the target rightward, whereas the closer right source pushes it leftward. Coulomb’s law states that force varies as 1/r²; therefore the smaller distance on the right produces the larger force. The stronger leftward force is not fully balanced by the weaker rightward force, so the net force points left. Option A is correct.

Like charges repel, so the closer negative charge on the left pushes the middle negative charge toward the right. The equal negative charge on the right repels it toward the left. Although the charge magnitudes are equal, the distances are not. Since F = k|q1q2|/r², the smaller left distance gives a stronger force than the right distance. The stronger contribution wins, leaving a rightward net force. Therefore option A is correct.

For equal source charges acting on the same target, Coulomb’s law gives F ∝ 1/r². Let the nearer and farther distances be 2x and 5x. Then F_near = kQq/(2x)² = kQq/(4x²), while F_far = kQq/(5x)² = kQq/(25x²). Hence F_near/F_far = (1/4)/(1/25) = 25/4. The required ratio is twenty-five to four, so option C is correct.

For each source charge, F = k|q_target Q|/r², so the force ratio is (Q1/Q2)(r2/r1)². Using Q1/Q2 = 16 and r1/r2 = 4 gives F1/F2 = 16 × (1/4)² = 16 × 1/16 = 1. The sixteenfold charge advantage is exactly cancelled by the sixteenfold reduction caused by four times the distance. Therefore the forces are equal, and option C is correct.

Coulomb’s law gives F ∝ Q/r² when the target charge is fixed. If the distance changes from r to 2r, the force from the same source charge becomes Q/(2r)² = Q/(4r²), which is one-fourth of its original value. To restore the original force, the source charge must be multiplied by four: (4Q)/(2r)² = Q/r². Hence option B, four times, is correct.

With the target charge fixed, Coulomb’s law requires F ∝ Q/r². Increasing the distance from r to 5r reduces the force of an unchanged source charge by (1/5)² = 1/25. To compensate, the source-charge magnitude must increase by 25. Indeed, the new force is k(25Q)q/(5r)² = 25kQq/(25r²) = kQq/r², equal to the original force. Thus option C is correct.

According to Coulomb’s law, unlike charges attract and like charges repel. If the original charges are +q and −Q, changing both signs gives −q and +Q. Their signs are still opposite, so the product of the charges remains positive in magnitude but the interaction remains attractive. The force magnitude is also unchanged because the charge magnitudes and separation are unchanged; only the algebraic signs have both reversed. Thus option A is correct, not B, C, or D.

Repulsion shows that the two original charges have the same sign, such as +q and +Q or −q and −Q. Reversing the sign of only one charge changes the pair into unlike charges, for example +q and −Q. Coulomb’s law states that unlike charges attract. The magnitudes and separation have not been specified to change, so the force does not become zero and the distance is not automatically altered. Therefore option B is correct.

For each source charge, Coulomb’s law gives Fᵢ = k|qᵢQ|/rᵢ², where Q is the target charge. Thus every individual force on the target is directly proportional to |Q|. Replacing Q by 3Q makes every force vector three times as large. Since all directions and source charges remain unchanged, their vector sum also scales by three: F_net' = 3F_net. Therefore option A is correct; it is not nine times because only one charge is changed.

The magnitude of the Coulomb force between a source charge q and target charge Q is F = k|qQ|/r². This expression uses the absolute values of the charges, so reversing Q to −Q or −Q to +Q does not change |Q| and therefore does not change F. The direction or nature of the interaction may reverse—from attraction to repulsion or vice versa—but the magnitude remains unchanged when distance and all other magnitudes stay fixed. Hence option A is correct.

The net force on a specified target is found by adding, as vectors, only the forces whose point of application is that target. A force between two external charges acts on those two external charges, not directly on the target. It may influence their motion and therefore a later configuration, but it is not one of the instantaneous force contributions on the target. All electrostatic forces still have the same unit, newton, so option A is the correct reason.

For two forces of magnitude F separated by angle θ, the resultant magnitude is R = √(F² + F² + 2F²cosθ). At θ = 180°, cosθ = −1, so R = √(2F² − 2F²) = 0. Equivalently, the two equal vectors point in exactly opposite directions and cancel pairwise. Therefore the net force is zero. Options B, C, and D would require unequal magnitudes or a different angle, so option A is correct.

An angle of 0° means that the two force vectors point in the same direction. If each force has magnitude F, vector addition gives R = F + F = 2F. The general formula also confirms this: R = √(F² + F² + 2F²cos0°) = √(4F²) = 2F. Thus the resultant is twice either individual force. It is not zero or half, because cancellation occurs only for opposite directions. Option C is correct.

Let each perpendicular force have magnitude F. Because their directions are at 90°, the resultant is R = √(F² + F²) = F√2, approximately 1.414F. Hence it is greater than one individual force F but less than the arithmetic sum 2F. A resultant magnitude cannot be negative, and the statement that it is not greater than any force is false. Therefore option A is the only correct choice.

The problem directly states that the force in the medium is one-third of its value in air. Therefore F_medium = F_air/3 = 30 N/3 = 10 N. The dielectric properties of the medium are already represented by the given one-third factor, so no additional factor should be introduced. Ninety newtons would mean multiplication by three, thirty would mean no change, and fifteen would mean one-half. Hence option A is correct.

In a uniform dielectric medium, the Coulomb force is reduced according to the medium’s relative permittivity: F = (1/4πε₀εᵣ)|q₁q₂|/r². In a multiple-charge problem, each pairwise force involving the relevant charges must be calculated with the appropriate medium factor before the force vectors are added. The effect is not restricted to the largest charge, and forces do not become zero merely because they are reduced. Therefore option A is correct.

The governing concept is superposition of forces: the net force is the vector sum of all individual Coulomb forces, F_net = F₁ + F₂ + ... . A zero result means equal and suitably directed forces may cancel, although each force can be nonzero. Thus option A is correct. Option B is too strong, while C and D are unnecessary conditions; cancellation can occur with finite distance and nonzero charges.

Forces obey vector addition, so both magnitude and direction matter. For two equal forces F separated by 60°, the resultant is √(F² + F² + 2F²cos60°) = √3F, which is nonzero. Cancellation of equal forces requires an angle of 180°, not 60°. Therefore A is correct. B and C contradict the stated situation, and D is false because force is a vector.

Coulomb’s law gives F = kq₁q₂/r². Let the original force be 72 N. Halving q₁q₂ multiplies the force by 1/2, while doubling r multiplies it by 1/2² = 1/4. Hence F' = 72 × 1/2 × 1/4 = 72/8 = 9 N. Option A is correct; 18 N ignores the distance effect and 36 N ignores one factor.

Use Coulomb’s law, F = kq₁q₂/r². Increasing the charge product fourfold makes the force 4F. Halving the distance changes r² to r²/4, so it contributes another factor of 4. Therefore F' = 9 × 4 × 4 = 144 N. Option C is correct. The values 36 N and 72 N account for only part of the change, while 18 N is inconsistent with both increases.

With unchanged charges, Coulomb’s law reduces to F ∝ 1/r². If the new distance is r' = 3r, then F'/F = (r/r')² = (1/3)² = 1/9. Thus tripling the separation makes the force one ninth of its original value, so option A is correct. Making the distance one third would increase force ninefold, while making it ninefold would reduce force to 1/81.

For fixed charges, Coulomb’s law states F ∝ 1/r². To obtain F'/F = 25, choose a distance ratio satisfying (r/r')² = 25. Hence r/r' = 5 and r' = r/5. The separation must become one fifth, so option B is correct. Increasing distance would reduce the force, and making it 25 times larger would reduce force to 1/625 rather than increase it.

The superposition principle says the force on the target is the vector sum of the forces due to the two source charges. Because the source charges are equal, positive, and placed on the same side at equal distance, their forces on a target located in the corresponding geometry point in the same direction and have equal magnitude. They therefore add, giving twice one force; option A is correct. They do not cancel or leave only one force.

By Coulomb’s law, equal source-charge magnitudes at equal distances exert equal-magnitude forces on the target. Since both interactions are attractive, each force points toward its source charge. The sources lie on opposite sides, so the two equal vectors point in opposite directions and cancel: F_net = F − F = 0. Thus option A is correct. A left or right result would require unequal magnitudes or an asymmetry.