Coulomb's Law (force between multiple charges)

The net force follows vector superposition. Equal charges at equal distances produce equal-magnitude forces on the target. With repulsion, each force pushes the target away from its source: the source on the left pushes right, and the source on the right pushes left. These equal opposite vectors cancel, so F_net = F − F = 0. Option A is correct; a left or right result would require unequal charge magnitudes or distances.

The governing idea is that net force is calculated on a specified particle, called the target charge. First identify that charge and define its position; then determine the force exerted on it by each external charge, including direction, and add the vectors. Therefore option A is correct. The appearance of a diagram and the largest numerical choice are irrelevant, while assuming every force is zero is physically unjustified.

The governing concept is vector superposition of electric forces. A target charge may experience several nonzero Coulomb forces, while their vector sum is zero. For example, two equal forces in opposite directions cancel, giving zero net force even though both source charges remain present and exert forces. Thus option A is correct. Option B wrongly confuses zero resultant with zero individual forces; C and D have no basis in Coulomb’s law.

The governing principle is superposition: the net force on a selected charge equals the vector sum of the individual Coulomb forces exerted by all other charges. First identify the target, then use F = k|qQ|/r² for every source charge, determine attraction or repulsion and its direction, and finally add components or vectors. Therefore A is correct; B ignores direction, C omits force information, and D calculates forces not required on the target.

The governing ideas are Coulomb’s law and symmetry. The midpoint is 6 cm from each outer charge, and the two source charges are equal, so each force on the central positive charge has the same magnitude, proportional to Qq/(6 cm)². Because like charges repel, the left source pushes right and the right source pushes left. Therefore the individual forces are equal and opposite, making option A correct.

Each positive charge attracts the negative charge, so the left charge pulls it leftward and the right charge pulls it rightward. The charge magnitudes and distances are equal, so Coulomb’s law gives equal force magnitudes, proportional to Qq/r². These equal forces act in opposite directions and cancel vectorially. Hence the net force is zero, making option A correct; attraction does not mean the forces point in the same direction.

The principle of superposition says that the net force is the vector sum of all individual forces. Let each given force have magnitude F. Since both vectors point to the right, they are parallel and have the same direction, so their magnitudes add: F_net = F + F = 2F toward the right. Thus option C is correct. Opposite-direction subtraction would apply only if one force pointed left.

For perpendicular vectors, the Pythagorean theorem gives the resultant magnitude: R = √(F₁² + F₂²). Substituting the values, R = √(12² + 16²) = √(144 + 256) = √400 = 20 N. Therefore option B is correct. Simple subtraction gives 4 N, while direct addition gives 28 N; neither is valid for perpendicular forces, and multiplying the values is dimensionally inappropriate.

Represent the forces as rectangular vector components: F_x = +30 N because the horizontal component is rightward, and F_y = +40 N because the vertical component is upward. Both components are positive, so the resultant lies in the first quadrant, between the rightward and upward axes. Its magnitude would be √(30²+40²)=50 N, but only its quadrant is asked. Hence option A is correct.

The two forces are perpendicular, so their resultant is found using the right-triangle relation rather than ordinary addition. R = √(F_x² + F_y²) = √(15² + 20²) = √(225 + 400) = √625 = 25 N. Therefore option B is correct. The value 5 N comes from subtraction and 35 N from direct addition, neither of which applies to perpendicular components; 300 is not a force resultant.

The governing concept is vector addition of forces. The 8 N force to the right and the 8 N force to the left have equal magnitudes and opposite directions, so their horizontal resultant is 8 − 8 = 0 N. The remaining vertical component is 6 N upward. Hence the net force is 6 N upward, making option A correct; it is not zero because the upward force remains.

The principle is component-wise vector addition. The rightward 12 N and leftward 12 N forces are equal and opposite, so their horizontal components cancel: 12 − 12 = 0 N. The upward component is still 5 N, giving a resultant of 5 N upward. Therefore option A is correct. Option C incorrectly adds opposite horizontal forces, while option D ignores the remaining upward force.

Forces are vectors, so a zero resultant requires cancellation in every component. With only two forces, one force can cancel the other only when their magnitudes are equal and their directions are exactly opposite. Thus F₁ + F₂ = 0 implies F₁ = −F₂, making option A correct. Equal magnitudes in the same direction would add rather than cancel, and unequal forces cannot produce zero resultant by themselves.

Coulomb’s law gives F = k|q₁q₂|/r², so with charges fixed, force varies inversely as the square of distance. If F′ = 16F, then r′ must satisfy (r/r′)² = 16. Therefore r′ = r/4, meaning the distance must become one-fourth. Option B is correct; doubling or halving the distance would change the force by factors of 1/4 or 4, not 16.

According to Coulomb’s law, F is proportional to 1/r² when the charges are fixed. To obtain F′ = F/16, the square of the distance must increase by 16: (r′/r)² = 16. Taking the positive physical distance gives r′/r = 4, so the distance must be four times the original distance. Thus option A is correct; eight times would reduce the force to 1/64.

Coulomb’s law is F = kq₁q₂/r². Increasing one charge by a factor of 4 would increase the force by 4 if distance stayed fixed. To preserve the original force, r² must also increase by 4, so the distance must increase by √4 = 2. Therefore the distance must double, making option A correct. Keeping distance unchanged would leave the force four times larger.

Use Coulomb’s law, F = kq₁q₂/r². Multiplying one charge by 9 would multiply the force by 9 if r did not change. For the force to remain constant, the distance-squared factor must also increase by 9. Hence r′² = 9r² and r′ = 3r. Option A is correct; making the distance nine times would reduce the force excessively, to 1/9 of the original after the charge change.

The governing rule is vector addition. Equal magnitudes cancel only when the vectors point in opposite directions. Here both forces point in the same direction, so their magnitudes add: Fnet = F + F = 2F. Therefore option B is correct. Option A would apply to equal opposite forces, while option C would apply only if one force were absent or if another vector cancelled part of the sum.

For perpendicular force vectors, the resultant magnitude follows the Pythagorean theorem: Fnet = √(F₁² + F₂²). Substituting the values gives √(10² + 24²) = √(100 + 576) = √676 = 26 N. Thus option B is correct. Adding 10 and 24 directly would ignore the right angle, while multiplying them gives neither a force magnitude nor the correct vector resultant.

When two forces are perpendicular, their resultant is not found by ordinary addition; it is the hypotenuse of a right triangle. Therefore Fnet = √(7² + 24²) = √(49 + 576) = √625 = 25 N. Option B is correct. The sum 31 N would be valid only for forces in the same direction, and 168 N is merely their product, not a resultant magnitude.

For a charge in equilibrium, the vector equation is F1 + F2 + F3 = 0. If the first two forces have resultant R = F1 + F2, then F3 = −R. Thus, the third force must have the same magnitude as the resultant of the first two but point in the opposite direction. It need not be parallel to either individual force, so options B and C are not generally valid; option D ignores the equilibrium condition.

Coulomb’s law gives F = kq1q2/r². Let each original equal charge be q. After doubling the distance, let each charge become xq. The new force is k(xq)(xq)/(2r)² = x²kq²/(4r²), so Fnew/Fold = x²/4. Setting this ratio to 1 gives x = 2. Therefore each charge must be doubled; multiplying each by four would make the force four times too large.

Using Coulomb’s law, F = kq²/r² for two equal charges. If the distance becomes 3r and each charge becomes xq, then F′ = k(xq)²/(3r)² = x²F/9. Requiring F′ = F gives x² = 9, so the positive multiplication factor is x = 3. Thus each charge must be tripled. Making each charge nine times larger would increase the charge product by 81, not merely compensate for the distance change.

Attraction initially means the charges have opposite signs, so q1q2 is negative. Reversing only one sign makes the charges have the same sign, changing the interaction from attraction to repulsion. Doubling both magnitudes changes the absolute charge product from |q1q2| to |2q1·2q2| = 4|q1q2|. With distance unchanged, the repulsive force is therefore four times the original magnitude.

Initial repulsion means the two charges have the same sign. Reversing one sign makes their signs opposite, so the force becomes attractive. If each magnitude changes from q to q/2, the charge product becomes (q1q2)/4. Since the separation is unchanged, Coulomb’s law shows that the new force magnitude is one-fourth of the original, but its direction is attractive. Hence A is correct.