This problem tests electric charge and electric force. Initially, charges +2q and -q attract with force F = k(2q)(q)/(0.20)² = 2kq²/0.04. When separation becomes 0.10 m (halved), the new force is F' = 2kq²/(0.10)² = 2kq²/0.01 = 4F, making it four times as large. The force remains attractive since the charges have opposite signs. A common mistake is thinking halving distance doubles the force (answer C), forgetting the inverse-square relationship. Always square the distance ratio: (1/2)² = 1/4, so force increases by factor of 4.

This question tests understanding of electric charge and electric force. By Coulomb's law, the force between two charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. Since sphere 1 has charge +q and sphere 2 has charge -2q, the charges have opposite signs, making the force attractive. The magnitude is F = k|q₁||q₂|/r² = k|q||-2q|/r² = 2kq²/r². A common misconception is forgetting to use the absolute values when calculating magnitude, which might lead someone to think the answer involves negative values. Always determine force direction from charge signs first, then calculate magnitude using absolute values.

This question tests understanding of electric charge and electric force. Coulomb's law states that the electric force between two charges is inversely proportional to the square of the distance between them: F = kq₁q₂/r². When the distance doubles from r to 2r, the new force becomes F' = kq²/(2r)² = kq²/4r² = F/4. Since both charges are positive (+q), the force is repulsive, but the question asks only about magnitude. A common misconception is thinking force is inversely proportional to distance (not distance squared), which would incorrectly suggest the force becomes half as large. Always remember that electric force follows an inverse square law with distance.

This problem tests understanding of electric charge and electric force. Since both Q₁ (-q) and Q₂ (-2q) are negative charges, they have the same sign, which means they repel each other according to Coulomb's law. The force on Q₁ due to Q₂ points away from Q₂, along the line connecting the two charges. The fact that Q₂ has twice the magnitude of Q₁ affects the force strength but not the direction—repulsion is determined by the like signs. A common misconception is thinking that two negative charges somehow attract or that the force is zero, but like charges always repel regardless of their magnitudes. Always determine force direction by the charge signs: like charges repel, opposite charges attract.

This question tests understanding of electric charge and electric force. Using Coulomb's law F = kq₁q₂/r², the force between charges +2q and +q is F = k(2q)(q)/r² = 2kq²/r². Since both charges are positive, like charges repel, making the force repulsive and directed away from the +q charge. The force on the +2q sphere has the same magnitude as the force on the +q sphere by Newton's third law. Students who choose option A might forget that like charges repel, incorrectly thinking all electric forces are attractive. Always check the signs of both charges to determine whether the force is attractive or repulsive.

This question tests understanding of electric charge and electric force. Initially with two +q charges, the force is repulsive with magnitude F = kq²/r². When one charge changes to -q, we now have opposite charges (+q and -q), so the force becomes attractive. The magnitude remains F = k|q||q|/r² = kq²/r², which is the same as before. Only the direction changes from repulsive to attractive. Students who choose option C might think changing the sign somehow reduces the force magnitude, not realizing that Coulomb's law uses the absolute values of charges for magnitude. Always remember that changing charge signs affects force direction but not magnitude.

This problem tests understanding of electric charge and electric force. The original force between +2q and +q at distance r follows F₁ ∝ (2q)(q)/r². The new scenario asks about the force between +2q and the third charge +q at distance 2r, giving F₂ ∝ (2q)(q)/(2r)² = 2q²/4r² = (1/2)q²/r². Comparing to the original force, F₂ = (1/4)F₁. Both forces are repulsive since all charges are positive. Choice D (unchanged) is incorrect because it ignores that the distance has changed from r to 2r—failing to apply the inverse square law. Always identify which two charges and what separation distance you're analyzing.

This problem tests understanding of electric charge and electric force. The original force between +q and +4q is F₁ ∝ (q)(4q) = 4q². When +4q is replaced with +2q, the new force becomes F₂ ∝ (q)(2q) = 2q². Comparing these, F₂ = (2q²)/(4q²) × F₁ = (1/2)F₁. Both configurations produce repulsive forces since all charges are positive. Choice D (unchanged) is incorrect because it ignores that one charge magnitude has changed—perhaps assuming only separation affects force magnitude. Always account for changes in both charge magnitudes when comparing electric forces.

This problem tests understanding of electric charge and electric force. According to Coulomb's law, the force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them (F ∝ q₁q₂/r²). When the separation doubles from r to 2r, the denominator becomes (2r)² = 4r², making the force 1/4 as large. The attractive force between the opposite charges (+q and -2q) still points along the line joining them, but with reduced magnitude. Choice A (1/2 as large) is incorrect because it assumes force is inversely proportional to distance rather than distance squared—a common misconception. Always remember that electric force follows an inverse square law with distance.

This problem tests understanding of electric charge and electric force. Initially, two +q charges repel with force magnitude F. When the second sphere's charge changes from +q to -q, the charges now have opposite signs, so they attract instead of repel. According to Coulomb's law, the force magnitude remains F = kq²/d² because the absolute values of the charges haven't changed. However, the direction reverses: instead of pointing away from the first sphere (repulsion), the force now points toward the first sphere (attraction). A common misconception is thinking that changing the sign somehow affects the magnitude or that opposite charges cancel to give zero force. Always analyze both magnitude (from Coulomb's law) and direction (from charge signs) separately.