Coulomb's Law (force between multiple charges)

Let each original charge be q. Reducing both to q/6 changes their product from q² to q²/36, which would reduce the force to 1/36 at the same distance. Since F ∝ q₁q₂/r², the distance must be reduced so that 1/r² becomes 36 times larger. Taking r′ = r/6 gives exactly that compensation. Therefore option A is correct; increasing or keeping distance would reduce force further.

From Coulomb’s law, F′/F = (Q′/Q)(r/r′)². Here 9 = 81(r/r′)², so (r/r′)² = 1/9 and r′/r = 3. Equivalently, the 81-fold charge-product increase must be reduced by a factor of 9 through the inverse-square distance term. Therefore tripling the distance produces the required ninefold force.

Choose one corner charge as the target. The other two positive charges repel it with equal forces of 70 N, and the angle between those force vectors is 60°, the triangle’s interior angle. Hence R = √(70² + 70² + 2·70²cos60°) = 70√3 N. The forces do not cancel because cancellation requires equal opposite directions.

Use Coulomb’s inverse-square law and compare the two force magnitudes on the target. The left charge repels the positive target to the right with relative factor 50/5² = 50/25 = 2. The right charge repels it to the left with factor 8/2² = 8/4 = 2. The magnitudes are equal and directions opposite, so the vector sum is zero; option A is correct.

Coulomb’s law gives force proportional to the source-charge magnitude divided by the square of distance: F ∝ Q/r². The left contribution is 108/6² = 108/36 = 3 in relative units and points right because both charges repel. The right contribution is 27/3² = 27/9 = 3 and points left. Equal opposite forces cancel, so the net force is zero. Therefore, option C is correct; either directional option ignores the cancellation.

By Coulomb’s law, each force is proportional to |Q|/r². The left negative charge attracts the positive target leftward with relative factor 128/8² = 128/64 = 2. The right negative charge attracts it rightward with factor 32/4² = 32/16 = 2. These equal forces act in opposite directions, so their vector sum is zero. Thus option A is correct; neither directional option remains after cancellation.

Coulomb’s law gives F ∝ 1/(εr²) when charges are fixed. The medium changes the force factor to 1/16, while halving the distance multiplies the force by 1/(1/2)² = 4. Thus the combined factor is (1/16)×4 = 1/4. The new force is 720×1/4 = 180 N, so option B is correct. Ignoring either effect leads to the other numerical distractors.

Coulomb’s law gives F ∝ q1q2/(εr²). The medium changes the original force by the factor 1/64. Increasing one charge eightfold multiplies the force by 8, so the combined factor is (1/64) × 8 = 1/8. Therefore, Fnew = 384 × 1/8 = 48 N. Hence option B is correct; 6 N uses an incorrect factor, 64 N ignores the charge change, and 384 N ignores the medium.

For fixed medium, Coulomb’s law is F = k(q1q2)/r². If r becomes 17r, the denominator becomes (17r)² = 289r². To keep F unchanged, the charge product q1q2 must also be multiplied by 289. Thus the required product is 289 times its original value, so option C is correct. A factor of 17 overlooks the square dependence on distance, while 1/17 changes the force in the wrong direction.

Using F ∝ Q/r², the force ratio is Fnew/Fold = 72/(rnew/rold)². We require this ratio to be 8, so 72/x² = 8. Hence x² = 9 and x = 3, taking the positive distance factor. Therefore the separation must be tripled, making option A correct. Doubling gives a force factor 18, leaving the distance unchanged gives 72, and increasing it ninefold gives only 8/9 of the original force.

Coulomb’s proportional relation is F ∝ Q/r², where Q is the charge product. Let the distance factor be x. Then Fnew/Fold = (1/2)/x². The required ratio is 1/18, so (1/2)/x² = 1/18, giving x² = 9 and x = 3. Thus the separation must be tripled, which is option A. Doubling would give 1/8, not 1/18; keeping distance unchanged would give only 1/2.

Take the target charge as positive and compare force magnitudes using F ∝ |Q|/r². The left positive charge repels the target to the right with relative magnitude 125/5² = 125/25 = 5. The right negative charge attracts the positive target to the right with magnitude 200/10² = 200/100 = 2. Since both forces point rightward, their resultant also points rightward. Thus option A is correct; they do not oppose each other.

Both external charges are negative, so each attracts the positive target. The left charge pulls left with relative magnitude 256/8² = 256/64 = 4. The right charge pulls right with relative magnitude 64/4² = 64/16 = 4. These forces have equal magnitudes and opposite directions, so vector addition gives a zero net force. Therefore option A is correct; neither side dominates and the result is not twice left.

The target is negative and both external charges are positive, so both interactions are attractive. The upper charge pulls upward with relative magnitude 216/6² = 216/36 = 6. The lower charge pulls downward with magnitude 54/3² = 54/9 = 6. Equal forces in opposite vertical directions cancel exactly, giving zero resultant force. Hence option A is correct; an upward or downward answer would ignore the equality of the two force magnitudes.

The two outer charges repel the positive middle charge in opposite directions. Because the spacing from the middle charge is equal, each force is proportional only to the corresponding outer charge. The left force points right and has relative magnitude 1/6. The right force points left and has relative magnitude 18. Since 18 is much larger than 1/6, the resultant points left. Thus option A is correct; equal spacing does not make the forces equal after the charge changes.

Since all three charges are negative, the outer charges repel the negative middle charge. The left charge pushes the middle charge to the right, while the right charge pushes it to the left. Equal spacing means the force comparison follows the modified charge factors: 12 for the left force and 1/8 for the right force. The rightward force is therefore dominant, so the net force is to the right. Option A is correct; the forces are not equal.

For two charges, Coulomb’s law is F ∝ q₁q₂/r². Reducing each charge to 1/7 makes the product q₁q₂ equal to 1/49 of its original value. To compensate and preserve the same force, r² must also become 1/49 of its original value. Hence r′ = r/7. The distance must be one seventh, so option A is correct; one forty-ninth is the factor for r², not r.

The electric dipole moment p is defined from the negative charge to the positive charge. On the axial line outside the dipole, the field contributions combine so that the far-point field is directed along p on the positive side of the dipole. Its magnitude varies approximately as E = (1/4πε₀)(2p/r³) for a far point. Therefore option A is correct; the opposite direction applies on the negative axial side, not universally.

For a dipole, the dipole moment points from the negative charge to the positive charge. At a point on the equatorial line, the components of the two charge fields perpendicular to the dipole axis cancel, while the components along the axis reinforce in the direction opposite to p. For a far point, E ≈ (1/4πε₀)(p/r³) in magnitude and is opposite to p. Hence option A is correct.

Consider equal charges +q and −q separated on one axis. Between them, the field due to +q points away from +q and the field due to −q points toward −q; both point in the same direction, so they add. Outside the pair, the fields oppose, but equality would require equal distances from both charges, which only occurs at the midpoint, not outside. At the midpoint the fields add, so no finite zero-field point exists on the axial line.

Gauss’s law states that the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. The external positive charge does produce an electric field on the surface, so option B is false. However, because the charge lies outside, field lines that enter the surface must also leave it. Their inward and outward contributions cancel, while the enclosed charge is zero; therefore the net flux is zero. Hence option A is correct.

For a dipole in a uniform electric field, the torque is τ = pE sin θ and acts to reduce the angular displacement from the field direction when the dipole is near θ = 0. At exact alignment the torque is zero, but this is stable equilibrium. After a small rotation, the torque acts oppositely to the displacement and restores alignment. Thus option A is correct; option B describes unstable behavior.