Coulomb's Law (force between multiple charges)

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: the forces from that pair have equal magnitudes and opposite directions, regardless of whether the pair is positive or negative. Each pair therefore cancels by symmetry. Since both opposite pairs cancel separately, the vector sum of all four forces is zero. Option A is correct.

The governing idea is the principle of superposition: the net force on a chosen target charge equals the vector sum of the individual Coulomb forces exerted by all other charges. For each source charge, calculate k|qQ|/r² and determine whether the force is attractive or repulsive. Resolve components when needed, then add them. Therefore A is correct; B ignores directions, C omits forces, and D adds forces that do not act on the target.

For two equal forces F inclined at angle θ, the resultant is R = √(F² + F² + 2F²cosθ). With θ = 60° and cos60° = 1/2, R = √(3F²) = F√3. Given R = 30√3 N, dividing by √3 gives F = 30 N. Therefore option C is correct; substituting 10, 15, or 60 N would not produce the stated resultant.

The chosen charge is repelled by the other two positive charges, so two forces of 50 N act along the two sides meeting at its corner. The angle between them is 60°. Using R = √(F² + F² + 2F²cos60°), R = √(3F²) = 50√3 N. Thus C is correct. The forces do not cancel because their directions are separated by 60°, not 180°.

Both source charges are positive, so each repels the positive target. The left charge pushes it right with a proportional factor 18/3² = 18/9 = 2. The right charge pushes it left with factor 8/2² = 8/4 = 2. The magnitudes are equal and directions opposite, so the vector sum is zero. Hence A is correct; the other directional choices ignore this exact balance.

Using F = kQ/r², the charge-product change alone would multiply the force by 20. The required overall factor is only 5, so the distance effect must contribute a factor of 5/20 = 1/4. Because force varies inversely as the square of distance, replacing r by 2r makes the force one-fourth. Therefore the separation must be doubled, and option A is correct. Halving the distance would instead increase the force by four.

Coulomb’s law gives F ∝ Q/r². Halving the charge product first changes the force by a factor of 1/2. To obtain the required final factor 1/8, the distance must contribute (1/8)/(1/2) = 1/4. A distance increase from r to 2r makes the inverse-square factor 1/4. The combined factor is therefore (1/2)(1/4) = 1/8, so option A is correct.

Take the target as positive and compare force factors using |q|/r². The positive charge on the left repels the target to the right, with factor 27/3² = 27/9 = 3. The negative charge on the right attracts the positive target toward the right, with factor 72/6² = 72/36 = 2. Both forces point rightward, so their resultant also points rightward. Option A is correct; the magnitudes need not be equal because their directions are the same.

Opposite charges attract. Therefore, the negative charge on the left pulls the positive target leftward, with relative factor 50/5² = 50/25 = 2. The negative charge on the right pulls it rightward, with relative factor 8/2² = 8/4 = 2. These forces have equal magnitudes but opposite directions, so they cancel vectorially. The net force is zero, making option A correct; neither side dominates.

A negative target is attracted toward each positive external charge. The upper positive charge therefore pulls upward, with relative factor 32/4² = 32/16 = 2. The lower positive charge pulls downward, with factor 8/2² = 8/4 = 2. Since these equal forces act along one line in opposite directions, their vector sum is zero. Thus option A is correct, not an upward or downward resultant.

The outer charges and the middle charge are all positive, so both outer charges repel the middle one. The left charge pushes the middle charge to the right, while the right charge pushes it to the left. Their distances from the middle are equal, so force magnitudes are proportional to their charge multipliers: 8 on the left and 2 on the right. The rightward force is larger, leaving a net force toward the right. Thus option A is correct.

Like charges repel, so the left negative charge pushes the middle negative charge to the right, while the right negative charge pushes it to the left. The two distances are equal, meaning the force magnitudes are proportional to the outer-charge multipliers. The leftward force from the right charge has factor 9, whereas the rightward force from the left charge has factor 2. Since 9 is larger than 2, the resultant is leftward. Therefore option A is correct.

Coulomb’s law is F = kq₁q₂/r². Reducing each charge to one-fifth makes the product q₁q₂ equal to 1/25 of its original value. To keep F unchanged, 1/r² must become 25 times larger. Thus r² must become 1/25 of its original value, giving r′ = r/5. Therefore option A is correct; changing distance by five times in the opposite direction would not compensate.

All four corners are at the same distance from the centre, so each force has the same magnitude because the charge magnitudes are equal. Charges at opposite corners produce forces along the same diagonal but in opposite directions: each opposite pair cancels exactly, regardless of whether the target charge is positive or negative. Both pairs therefore cancel, giving zero net force. Option A is correct.

By Coulomb’s law, the force magnitude is proportional to the source charge divided by the square of its distance. The left positive charge repels the positive target to the right, with relative factor 125/5² = 5. The right positive charge repels it to the left, with factor 28/3² = 28/9, about 3.11. Since 5 is larger, the resultant force is rightward. Thus option A is correct; zero would require equal opposite forces.

The left charge is negative like the target, so it repels the target toward the right. Its relative force factor is 64/4² = 64/16 = 4. The right charge is positive and attracts the negative target toward the right; its factor is 1/2² = 1/4. Since both forces point rightward, their magnitudes add rather than cancel. The net force is therefore to the right, so option A is correct.

For two forces of equal magnitude F separated by angle θ, the resultant is R = √(F² + F² + 2F²cosθ) = 2F cos(θ/2) over this interval. As θ increases from 0° to 180°, cos(θ/2) decreases from 1 to 0. Therefore the resultant is smallest at θ = 180°, where equal opposite forces cancel and R = 0. Option C is correct.

The stated medium effect first changes the air force to 450/9 = 50 N. Coulomb’s law gives F ∝ 1/r², so halving the separation multiplies the force by (1/(1/2)²) = 4. Therefore the final force is 50 × 4 = 200 N, equivalently 450 × (1/9) × 4 = 200 N. Option C is correct; 50 N ignores the distance change.

Using F ∝ q₁q₂/r², the charge-product change alone would multiply the force by 28. The required final factor is only 7, so the distance factor must reduce the force by 7/28 = 1/4. Since force varies inversely as r², making the distance 2r changes the force by 1/2² = 1/4. Therefore the distance must be doubled, making option A correct.

Coulomb’s relation gives F ∝ (q₁q₂)/r². Changing the charge product to 2/5 initially changes the force by 2/5 if distance is fixed. The desired factor is 1/10, so the distance factor must be (1/10)/(2/5) = 1/4. Because force varies as 1/r², a factor of 1/4 is obtained by doubling the distance: 1/(2²) = 1/4. Therefore option A is correct.

Take the target charge and the common Coulomb constant as reference. The left positive charge repels the positive target to the right with relative magnitude 64/4² = 64/16 = 4. The right negative charge attracts the target to the right with relative magnitude 128/8² = 128/64 = 2. Both forces point rightward, so their resultant is rightward. Option A is correct; zero and leftward would require cancellation or an opposite force direction.

Both external charges are negative, so each attracts the positive target. The left charge produces a leftward force proportional to 108/6² = 108/36 = 3. The right charge produces a rightward force proportional to 27/3² = 27/9 = 3. These equal forces act in opposite directions and cancel exactly. Therefore the net force is zero, making option A correct; neither side dominates.

The negative target is attracted toward each positive charge. Thus the upper charge pulls it upward with relative magnitude 75/5² = 75/25 = 3, while the lower charge pulls it downward with relative magnitude 27/3² = 27/9 = 3. Equal upward and downward forces cancel, giving zero resultant force. Hence option A is correct; an upward or downward answer would incorrectly ignore one equal contribution.

Let F be the force that an unchanged outer charge would exert at the common separation. Since all charges have the same sign, the left charge repels the middle charge rightward with magnitude F/4. The right charge repels it leftward with magnitude 12F. The leftward force is larger, and the net magnitude is 12F − F/4 = 11.75F. Therefore the net force points left, so option A is correct.

Like charges repel, so the enlarged left negative charge pushes the middle charge to the right. Its force is 9F if F is the original equal-charge force at that separation. The reduced right negative charge pushes the middle charge to the left with magnitude F/5. Since 9F is much larger than F/5, the resultant is rightward, with magnitude 9F − F/5 = 8.8F. Thus option A is correct.