Coulomb's Law (force between multiple charges)

The eastward and northward forces are perpendicular components of one resultant. Therefore the Pythagorean relation applies: R = √(F₁² + F₂²) = √(16² + 30²) = √(256 + 900) = √1156 = 34 N. Hence option C is correct. Adding the components directly would give an incorrect value because they are perpendicular, while fourteen and forty-six do not satisfy the vector-magnitude calculation.

Treat west as the negative horizontal direction and north as the positive vertical direction. The resultant has a westward component of 40 N and a northward component of 9 N, so it must lie between west and north. Its magnitude would be √(40² + 9²) = 41 N, confirming that both components contribute. Therefore the direction is north-west, option A. East or south components are absent, ruling out the other three quadrants.

The rightward and upward forces are perpendicular, so their resultant follows the Pythagorean theorem. Calculate R = √(28² + 21²) = √(784 + 441) = √1225 = 35 N. Thus option B is correct. Seven is the difference of the components and is not the resultant magnitude; forty-nine and 588 do not result from the required square-root calculation.

For two equal forces F separated by angle θ, the resultant magnitude is R = √(F² + F² + 2F²cosθ) = 2F cos(θ/2). For nonzero F, R can be zero only when cos(θ/2) = 0, which requires θ = 180°. Thus the forces must be equal, collinear, and opposite. Option A is correct; the same direction gives 2F, perpendicular forces give √2F, and an arbitrary angle does not cancel them.

Coulomb’s law gives F proportional to q1q2 and inversely proportional to the medium’s permittivity factor and r². The medium reduces the original force by 9, so its factor is 1/9. Tripling both charges multiplies q1q2 by 3 × 3 = 9. Hence the total factor is (1/9) × 9 = 1, and the new force remains 90 N. Therefore, option B is correct; option A ignores the charge change, while C applies only the charge factor.

Coulomb’s law is F = kq1q2/r². If r becomes 7r, the denominator becomes (7r)² = 49r², which would reduce the force to 1/49 of its former value if the charges were unchanged. To preserve the same force, q1q2 must also be multiplied by 49. Thus option C is correct. Seven times would compensate only a linear dependence, not the inverse-square dependence.

For a fixed external charge and separation, Coulomb’s law gives the force magnitude proportional to the magnitude of the target charge. Multiplying that magnitude by 5 therefore multiplies every individual force by 5. Reversing the target’s sign changes attraction into repulsion or repulsion into attraction, so the direction of each nonzero force reverses. Thus option A is correct; the force does not become zero or one-fifth.

All charges are positive, so each outer charge repels the positive middle charge. The left charge pushes the middle charge rightward, with a relative force proportional to 2/r². The right charge pushes it leftward, with a relative force proportional to 7/r². The distances are equal, so the common factors cancel in comparison; the leftward force is larger. Therefore option A is correct.

Like charges repel. Thus the left negative charge pushes the middle negative charge to the right, with a relative factor (1/3)/r². The right negative charge pushes it to the left, with factor 3/r². Equal spacing means the denominators are identical, so the rightward and leftward contributions compare as 1/3 to 3. The leftward force is greater, making option A correct.

The governing concept is vector addition. The eastward and westward forces lie on the same line and oppose each other, so their resultant is 35 − 11 = 24 N eastward. This 24 N component is perpendicular to the 7 N northward component. Therefore, F = √(24² + 7²) = √625 = 25 N. Thus option A is correct; simply adding all magnitudes would ignore direction.

Use vector addition by resolving forces along perpendicular directions. The northward and southward forces oppose one another, leaving 29 − 9 = 20 N northward. The westward component is 21 N and is perpendicular to this remainder. Hence F = √(20² + 21²) = √841 = 29 N. Option A is therefore correct. Adding 29, 9, and 21 directly would incorrectly treat opposite forces as parallel in the same direction.

For two equal forces F making an angle θ, the resultant is R = √(F² + F² + 2F²cosθ) = 2F cos(θ/2) for 0° ≤ θ ≤ 180°. At 0°, R = 2F, its maximum value; at 180°, R = 0 because the forces cancel. Since cos(θ/2) decreases throughout this interval, the resultant continuously decreases. Therefore option A is correct.

Coulomb’s law states that, for unchanged charges, the magnitude of force varies inversely as the square of separation: F ∝ 1/r². If the new force is 49F, then 49F/F = (r/r′)² = 49. Taking the positive square root gives r′/r = 1/7. Thus the distance must become one-seventh of its original value, so option B is correct; making it seven times would reduce the force instead.

For fixed charges, Coulomb’s law gives F = k|q₁q₂|/r², so F is inversely proportional to r². To obtain F′ = F/49, the squared distance must increase by 49: (r′/r)² = 49. Therefore r′/r = 7, taking the physically meaningful positive distance ratio. The separation must be seven times the original distance, making option A correct. A one-seventh distance would instead increase the force forty-nine times.

Coulomb’s law gives F ∝ q₁q₂/r². If one charge becomes 36 times larger while the other remains unchanged, the charge product and therefore the force would become 36 times larger at the old distance. To preserve the original force, the factor r² must also become 36. Thus r′ = √36 r = 6r. The distance must be six times the original, so option A is correct; keeping distance unchanged would not maintain the force.

Apply Coulomb’s law, F ∝ q₁q₂/r². Reducing each charge to one-fourth changes the charge product to (1/4)(1/4) = 1/16 of its original value. To keep F unchanged, the factor 1/r² must increase sixteenfold. This occurs when the new distance is r′ = r/4, because 1/(r/4)² = 16/r². Hence option A is correct; increasing the distance would reduce the force further.

The governing principle is superposition of vectors. Two forces that have equal magnitudes and exactly opposite directions add to zero: F₁ + F₂ = 0. The remaining total is therefore 0 + F₃ = F₃, regardless of the angle between F₃ and either cancelled force. Hence the net force has the same magnitude and direction as the third force, so option A is correct. Option C would apply only if the third force were also cancelled.

For two perpendicular force vectors, the resultant follows the Pythagorean theorem: F_net = √(F₁² + F₂²). Substituting the values gives F_net = √(20² + 21²) = √(400 + 441) = √841 = 29 N. Therefore option B is correct. The arithmetic sum, 41 N, would be valid only for forces in the same direction, while their difference, 1 N, would apply to opposite directions.

Because the two forces are perpendicular, their resultant is not found by ordinary addition or subtraction. Use F_net = √(F₁² + F₂²): F_net = √(18² + 24²) = √(324 + 576) = √900 = 30 N. Thus option B is correct. Six newton is the difference, 42 N is the arithmetic sum, and 432 is their product; none represents the perpendicular vector resultant.

The governing concept is vector equilibrium: the resultant of all forces is zero. Since the first two forces are perpendicular, their resultant is R = √(12² + 5²) = √169 = 13 N. The third force must be equal in magnitude and opposite in direction to this resultant, so its magnitude is 13 N. Seven N is the difference, while 17 N and 60 N do not follow the perpendicular-vector calculation.

Coulomb’s law gives F = kq₁q₂/r². Doubling the separation from 3 to 6 makes the force one-fourth if the charges remain unchanged. Therefore the product q₁q₂ must become four times larger to restore the original force. Because the charges are equal, changing each charge by a factor x changes their product by x²; x² = 4, so x = 2. Option A is correct.

Use Coulomb’s law, F = kq²/r², for equal charges. If the separation becomes r/3, the factor 1/r² makes the force nine times larger when q is unchanged. To compensate, q² must be reduced by a factor of nine. Hence each charge must be multiplied by 1/3, because (q/3)² = q²/9. Therefore option A is correct; one-ninth would reduce each charge too much.

Opposite charges attract, so the initial attraction means the two original charges had opposite signs. Reversing the sign of one charge makes both charges have the same sign, changing the interaction to repulsion. Coulomb’s law shows that force magnitude is proportional to |q₁q₂|. Multiplying one charge magnitude by six multiplies the force magnitude by six, while the unchanged distance keeps the distance factor constant. Thus option B is correct.

Repulsion initially means the two charges have the same sign. Changing the sign of only one makes their signs opposite, so the force becomes attractive. By Coulomb’s law, force magnitude is proportional to the product of the charge magnitudes. Reducing each charge to one-fourth changes the product to (1/4)(1/4) = 1/16 of its original value. Therefore the new force is one-sixteenth attractive, which is option A.

Forces on the positive target must be considered separately. The negative charge above attracts the target upward. The positive charge below repels the positive target, and that repulsion also pushes it upward, away from the lower charge. Because the distances and magnitudes are equal, the two forces have equal magnitudes and point in the same upward direction, so they add rather than cancel. Option A is correct.