Coulomb's Law (force between multiple charges)

For electrostatic interaction, F is proportional to q₁q₂/r². Repulsion initially means the charges have the same sign. Reversing only one sign makes the charges opposite, so the force changes to attraction. Reducing each magnitude to one-fifth changes the charge product by (1/5)(1/5) = 1/25. Therefore the new force is one twenty-fifth of the original and attractive: option A.

The force on a positive target due to a negative charge is attractive, so the upper negative charge pulls it upward. The lower positive charge has the same sign as the target and therefore repels it away from below, also upward. Since the distances and magnitudes are equal, the two upward forces have equal magnitude and add. Thus the net force is upward, option A.

Electrostatic force follows attraction between unlike signs and repulsion between like signs. The positive charge above attracts the negative target upward. The negative charge below repels the negative target away from itself, which is also upward. Equal distances and equal magnitudes make the two contributions equal, and they reinforce rather than cancel. Therefore the net force is upward, so option A is correct.

Coulomb’s law states that force varies inversely as the square of separation: F ∝ 1/r². If the side length is a, an adjacent corner is at distance a, while the opposite corner is at the diagonal distance √2a. Thus F_opposite/F_adjacent = a²/(√2a)² = 1/2. The opposite-corner force is therefore half the adjacent-corner force, making option A correct.

The two outer charges are at equal distances, so initially their forces on the middle positive charge are equal and opposite. The left positive charge repels the middle charge to the right, while the right positive charge repels it to the left. Reducing the left charge to one-eighth reduces its force to one-eighth, but the rightward force still comes from the weakened left charge. Therefore the stronger leftward force from the right charge remains, so option A is correct.

A negative charge is attracted toward each positive outer charge. Because both outer charges are equally distant, the force comparison depends only on their charge magnitudes. The left attraction has relative factor 4, whereas the right attraction has factor 1/3. These forces act in opposite directions, but 4 is much greater than 1/3, so the resultant points toward the left charge. Hence option A is correct.

The positive target is attracted toward both negative charges. The left charge produces a leftward force proportional to 72/6² = 72/36 = 2. The right charge produces a rightward force proportional to 18/3² = 18/9 = 2. These equal forces act in opposite directions, so they cancel exactly. The net force is therefore zero, making option A correct rather than either directional option.

East and north are perpendicular directions, so the resultant magnitude follows the Pythagorean theorem rather than ordinary addition. Thus R = √(39² + 80²) = √(1521 + 6400) = √7921 = 89 N. The value 119 N is the direct sum and ignores perpendicular geometry, while 41 N and 3120 N do not satisfy the vector calculation. Therefore option B is correct.

Take east as positive horizontal and north as positive vertical. The 63 N westward force gives a negative horizontal component, while the 16 N northward force gives a positive vertical component. Their vector sum therefore has west and north components simultaneously, placing it in the north-west quadrant. Its exact angle is not required to identify the quadrant. Hence option A is correct; the other choices have an incorrect horizontal or vertical sign.

The rightward and upward forces are perpendicular, so their resultant is the hypotenuse of a right triangle. Using vector addition, R = √(60² + 80²) = √(3600 + 6400) = √10000 = 100 N. The answer is not 140 N because perpendicular vectors are not added arithmetically, and 4800 N is their product. Thus option B is correct.

Coulomb’s law gives F proportional to the product of the charges and inversely proportional to the dielectric effect and distance squared. The medium first changes the force by a factor of 1/16. Increasing one charge fourfold multiplies the force by 4, so the combined factor is 4/16 = 1/4. Therefore, F_new = 144 × 1/4 = 36 N. Hence option B is correct; 9 N ignores the fourfold charge increase, while 144 N ignores the medium effect.

For fixed medium, Coulomb’s law is F = kq₁q₂/r². If the distance changes to 13r, the denominator becomes (13r)² = 169r². To keep F unchanged, the charge product q₁q₂ must also be multiplied by 169. Thus the required product is 169 times its original value, so option C is correct. Option A forgets the square dependence on distance, and option D gives the opposite trend.

For a fixed external charge and separation, Coulomb’s force is directly proportional to the target charge: F ∝ q_target. Increasing its magnitude by 11 multiplies the magnitude of every individual force by 11. Reversing the target’s sign changes attraction into repulsion or repulsion into attraction, so each force vector reverses direction. Therefore option A is correct; no cancellation or zero force follows automatically.

The governing concept is vector addition of forces. The opposite eastward and westward forces first combine to give 80 − 48 = 32 N eastward. This 32 N component is perpendicular to the 24 N northward component, so the resultant is √(32² + 24²) = √1600 = 40 N. Thus option A is correct; 32 N ignores the northward force, while 56 N and 104 N do not follow the perpendicular-component rule.

Use vector addition by resolving opposite directions first. The north–south component is 100 − 36 = 64 N northward. The remaining 48 N westward component is perpendicular to it. Therefore, F_net = √(64² + 48²) = √6400 = 80 N. Option A is correct. The value 64 N omits the westward force, while 112 N and 184 N are simple, invalid additions of components.

Coulomb’s law gives F = k|q₁q₂|/r². With the charges fixed, F is inversely proportional to r². If F′ = 169F, then r′² = r²/169, so r′ = r/13. Hence the distance must become one-thirteenth of its original value, making option B correct. Increasing distance would reduce force, so options A and C have the wrong trend; option D changes it excessively.

Coulomb’s law states that F ∝ 1/r² when the charges are fixed. To make the force 1/256 of its initial value, the squared distance must become 256 times larger. Since √256 = 16, the new distance must be 16r. Thus option A is correct. A shorter distance would increase the force, and changing the distance by 256 times would reduce the force by 256², not by 256.

Coulomb’s law is F = kq₁q₂/r². Increasing one charge by 64 multiplies the force by 64 if the distance is unchanged. To cancel that increase, r² must also increase by 64, so r must increase by √64 = 8. Therefore the distance should become eight times its original value, making option A correct. Keeping or reducing the distance would increase the force further.

For perpendicular force components, the magnitude of the resultant follows the Pythagorean theorem: F_net = √(56² + 90²). This equals √(3136 + 8100) = √11236 = 106 N. Hence option C is correct. Subtracting gives 34 N, adding gives 146 N, and multiplying gives 5040 N; none is valid for perpendicular vectors.

Because the two forces are perpendicular, their resultant is found from the right-triangle relation F_net = √(F₁² + F₂²). Substitution gives √(43² + 24²) = √(1849 + 576) = √2425 = 49.24 N, approximately 49 N. Therefore option B is the intended rounded answer. 19 N is the difference, 67 N is the direct sum, and 1032 N is the product.

Coulomb’s law gives F = kq₁q₂/r². When the distance changes from 5 to 10 units, it doubles, so the distance-squared factor becomes four times larger. To keep F unchanged, the product q₁q₂ must also become four times larger. Because the charges are equal and each is changed by the same factor x, their product changes by x². Thus x² = 4 and x = 2, so each charge must be doubled.

Coulomb’s law states that F is proportional to q₁q₂/r². If the distance becomes r/6, then the factor 1/r² makes the force 36 times larger when the charges are unchanged. To restore the original force, the product of the charges must therefore become 1/36 of its original value. If both equal charges are changed by the same factor x, then x² = 1/36, giving x = 1/6. Hence each charge becomes one-sixth.

Coulomb’s law determines the magnitude through |F| = k|q₁q₂|/r², while the signs determine whether the force is attractive or repulsive. Initial attraction means the charges have opposite signs. Reversing one sign makes their signs alike, so the force becomes repulsive. Multiplying one charge magnitude by 12 multiplies |q₁q₂|, and therefore the force magnitude, by 12, assuming the distance is unchanged. Thus the new force is 12 times repulsive.

For fixed separation, Coulomb’s law says the force magnitude is proportional to |q₁q₂|. Initially, repulsion means the charges have the same sign. Reversing only one sign makes the charges opposite in sign, so the interaction changes to attraction. Reducing each magnitude to one-sixth changes the product by (1/6)(1/6) = 1/36. Therefore the new force is one-thirty-sixth of the original magnitude and attractive.

The electric force direction follows attraction between unlike charges and repulsion between like charges. The negative charge above attracts the positive target upward. The positive charge below repels the positive target away from itself, which is also upward. Because the two source charges have equal magnitudes and equal distances, the two force magnitudes are equal; since both vectors point upward, they add rather than cancel. Therefore the net force is upward.