Coulomb's Law (force between multiple charges)

Use the sign rule for Coulomb force: unlike charges attract and like charges repel. The positive charge above attracts the negative target upward. The negative charge below repels the negative target away from the lower charge, which is upward as well. Equal distances and equal source-charge magnitudes make the two force magnitudes equal, but their directions coincide, so they reinforce one another. Hence the resultant force points upward, not downward or zero.

For point charges, Coulomb’s law gives F = kq²/r². Let the square side be a. An adjacent corner is at distance a, whereas the opposite corner is along the diagonal at distance √2a. Therefore F_opposite/F_adjacent = a²/(√2a)² = a²/(2a²) = 1/2. The opposite-corner force is consequently half the force from an adjacent corner. This compares individual force magnitudes, not the full vector sum from all three other corners.

All four corner charges are at the same distance from the centre, so equal charge magnitudes produce equal force magnitudes on the target. Consider each pair of opposite corners separately. The two forces from a pair have equal magnitudes and opposite directions along one diagonal, so they cancel. This cancellation occurs for the positive pair and independently for the negative pair, regardless of the target charge sign. Hence the vector sum of all four forces is zero.

Coulomb’s law gives F = kq1q2/r². Multiplying one charge by 9 multiplies the force by 9, while multiplying distance by 6 divides it by 6² = 36. Thus F′/F = 9/36 = 1/4, so F′ = 324/4 = 81 N. Option A would incorrectly use a factor of 6 rather than 36 for the distance change.

Use F = kq1q2/r². Increasing both charges by 5 makes their product 5 × 5 = 25 times larger. Increasing the distance by 10 makes the denominator 10² = 100 times larger. Therefore F′/F = 25/100 = 1/4, and F′ = 250/4 = 62.5 N. The other choices ignore one of these multiplicative effects or reverse it.

Coulomb’s law is F = k(q₁q₂)/r². For the force to remain constant, the ratio of charge product to r² must remain constant. If the charge product becomes 144 times larger, r′² must also become 144 times larger. Taking the positive square root gives r′ = √144 r = 12r. Therefore option B is correct; six times would provide only a 36-fold distance-squared increase.

The governing vector-addition relation for two equal forces P at angle θ is R = √(P² + P² + 2P²cosθ). For θ = 60°, R = √(3P²) = P√3. Since R = 50√3 N, division by √3 gives P = 50 N. The resultant itself is not one component, so option D is incorrect.

For two equal forces P acting at right angles, the vector-addition rule gives R² = P² + P² because the included angle is 90°. Hence R = P√2. Substituting the given resultant, 50√2 = P√2, so division by √2 gives P = 50 N. Option A would produce only 25√2 N, while options C and D would produce different resultants. Therefore option B is correct.

The requested resultant includes only the two specified adjacent-corner forces. Since each has magnitude 30 N and their directions are perpendicular, apply the Pythagorean relation: R = √(30² + 30²) = √1800 = 30√2 N. Adding them arithmetically would give 60 N, but that ignores their 90° angle; the forces also cannot cancel.

Coulomb’s law states that the force is proportional to source charge when the target charge and distance are unchanged. Initially, equal positive outer charges exert equal opposite repulsions on the middle positive charge. The left charge still produces the original rightward force, while the right charge produces only one ninth of the original leftward force. Hence the rightward force is larger and the resultant points right. Option A is correct; zero would require equal charges.

The negative target is attracted toward each positive source. Because both sources are at the same distance, their force ratio equals their charge ratio. The left attraction has relative magnitude 5, directed left, whereas the right attraction has relative magnitude 1/4, directed right. The leftward contribution is therefore much larger, so the resultant is leftward. Option A is correct; the force is not zero because the modified source charges are unequal.

The left source and the negative target repel, so the left source pushes the target rightward. Its relative force factor is 125/5² = 125/25 = 5. The positive source on the right attracts the negative target toward itself, also to the right. Thus both forces point right, regardless of the exact magnitude of the second force, and the net force must be rightward. Option A is correct; cancellation is impossible when the directions agree.

The eastward and northward forces are perpendicular components, so the resultant magnitude follows the Pythagorean relation R = √(Fx² + Fy²). Substituting gives R = √(65² + 72²) = √(4225 + 5184) = √9409 = 97 N. Therefore option C is correct. Subtraction gives 7 N, while direct addition would incorrectly ignore the perpendicular directions.

The rightward and upward forces are perpendicular, so their resultant is found from R = √(Fx² + Fy²). Here R = √(72² + 54²) = √(5184 + 2916) = √8100 = 90 N. Thus option B is correct. The difference 18 N is not the perpendicular resultant, 126 N is simple addition, and 3888 N is the product, none of which represents vector magnitude.

For two forces of equal magnitude F separated by angle θ, the resultant is R = √(F² + F² + 2F²cosθ) = 2F cos(θ/2) for 0°≤θ≤180°. This decreases as θ increases because cos(θ/2) decreases from 1 to 0. At θ = 0°, both forces act in the same direction and R = 2F, the maximum. Therefore option A is correct; at 180° they cancel.

For a fixed external charge and separation, Coulomb’s law gives F = k|qQ|/r², so the force magnitude is directly proportional to the target-charge magnitude. Multiplying that magnitude by 13 multiplies every individual force by 13. Reversing the target’s sign changes attraction into repulsion or repulsion into attraction, so the direction of each force reverses. Therefore option A is correct; no force becomes automatically zero.

The governing concept is vector addition of forces. The eastward and westward forces oppose one another, so their resultant is 120 − 68 = 52 N eastward. This 52 N component is perpendicular to the 39 N northward component. Therefore, F = √(52² + 39²) = √4225 = 65 N. Thus option A is correct; 52 N ignores the perpendicular component, while 91 N and 159 N do not follow the right-triangle relation.

Use vector addition by first combining opposite directions. The north–south component is 150 − 88 = 62 N northward. It is perpendicular to the 105 N westward component, so the net magnitude is F = √(62² + 105²) = √14929 = 122.2 N, which rounds to 122 N. Hence option A is correct. Option C gives only the vertical remainder, and option D incorrectly adds magnitudes directly.

Coulomb’s law gives F = k|q₁q₂|/r². With both charges fixed, F is inversely proportional to r². If the new force is 289F, then 289 = (r/r′)², so r′/r = 1/√289 = 1/17. The distance must therefore become one seventeenth of its original value, making option B correct. Increasing distance would reduce the force instead.

According to Coulomb’s law, F ∝ 1/r² when the charges do not change. To reduce the force to F/361, the square of the distance must increase by 361. Thus (r′/r)² = 361, giving r′/r = √361 = 19. The distance must become 19 times larger, so option A is correct. A factor of 361 applies to r², not directly to r; reducing the distance would increase the force.

Coulomb’s law states F ∝ q₁q₂/r². Increasing one charge by 81 multiplies the numerator by 81. To keep F unchanged, r² must also increase by 81, so the distance must increase by √81 = 9. Therefore the new distance is nine times the original distance, making option A correct. Making it 81 times would reduce the force far too much, while shortening or keeping it unchanged would increase the force.

For perpendicular force vectors, the resultant follows the Pythagorean theorem: Fnet = √(F₁² + F₂²). Substituting the values gives Fnet = √(72² + 95²) = √(5184 + 9025) = √14209 = 119 N. Therefore option C is correct. Subtraction gives 23 N, direct addition gives 167 N, and multiplying the magnitudes does not represent vector addition.

Because the two forces are perpendicular, their magnitudes cannot be added or subtracted directly. Apply the right-angle vector rule: Fnet = √(20² + 99²) = √(400 + 9801) = √10201 = 101 N. Thus option B is correct. The value 79 N is the difference, 119 N is the ordinary sum, and 1980 N is their product; none gives the perpendicular resultant.

Coulomb’s law gives F = kq₁q₂/r². The distance changes from 6 to 12, so it doubles and the factor r² becomes four times larger. To keep F unchanged, the product q₁q₂ must also become four times larger. Because the charges are equal and both are changed by the same factor x, their product changes by x². Thus x² = 4, giving x = 2. Therefore each charge must become twice its original value.

From Coulomb’s law, F ∝ q₁q₂/r². If the distance becomes r/8, then r² becomes r²/64, so the force would become 64 times larger if the charges stayed unchanged. To preserve the original force, the product of the two charges must therefore become 1/64 of its original value. Since the charges are equal and both are scaled by x, x² = 1/64, giving x = 1/8. Hence each charge should be one-eighth.