Study Apply Coulombs Law in Physics with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
All flashcards Flashcard 1: Find q 2 q_2 q 2 if F = 0.90 N F=0.90\ \text{N} F = 0.90 N , q 1 = + 1 × 10 − 6 C q_1=+1\times 10^{-6}\ \text{C} q 1 = + 1 × 1 0 − 6 C , and r = 0.10 m r=0.10\ \text{m} r = 0.10 m . Answer: ∣ q 2 ∣ ≈ 1.0 × 10 − 6 C |q_2|\approx 1.0\times 10^{-6}\ \text{C} ∣ q 2 ∣ ≈ 1.0 × 1 0 − 6 C . ∣ q 2 ∣ = F r 2 k ∣ q 1 ∣ = 0.90 × 0.01 8.99 × 10 9 × 10 − 6 ≈ 1.0 × 10 − 6 C |q_2| = \frac{Fr^2}{k|q_1|} = \frac{0.90 \times 0.01}{8.99\times10^9 \times 10^{-6}} \approx 1.0\times10^{-6}\ \text{C} ∣ q 2 ∣ = k ∣ q 1 ∣ F r 2 = 8.99 × 1 0 9 × 1 0 − 6 0.90 × 0.01 ≈ 1.0 × 1 0 − 6 C
Flashcard 2: What is the direction of the force between two charges with the same sign? Answer: Repulsive; along the line joining the charges. Like charges repel each other directly along their connecting line.
Flashcard 3: State the inverse-square relationship between force and separation distance in Coulomb's law. Answer: F ∝ 1 r 2 F \propto \frac{1}{r^2} F ∝ r 2 1 . Force decreases with the square of the distance between charges.
Flashcard 4: Calculate F F F for q 1 = + 1 × 10 − 6 C q_1=+1\times 10^{-6}\ \text{C} q 1 = + 1 × 1 0 − 6 C , q 2 = − 1 × 10 − 6 C q_2=-1\times 10^{-6}\ \text{C} q 2 = − 1 × 1 0 − 6 C , r = 0.10 m r=0.10\ \text{m} r = 0.10 m . Answer: F ≈ 0.90 N F\approx 0.90\ \text{N} F ≈ 0.90 N (attractive). F = 8.99 × 10 9 × 1 × 10 − 12 0.01 ≈ 0.90 N F = 8.99\times10^9 \times \frac{1\times10^{-12}}{0.01} \approx 0.90\ \text{N} F = 8.99 × 1 0 9 × 0.01 1 × 1 0 − 12 ≈ 0.90 N , opposite signs attract.
Flashcard 5: Calculate F net F_{\text{net}} F net on q 2 q_2 q 2 if q 1 = + 1 × 10 − 6 C q_1=+1\times10^{-6}\ \text{C} q 1 = + 1 × 1 0 − 6 C at 0.10 m 0.10\ \text{m} 0.10 m left and q 3 = + 1 × 10 − 6 C q_3=+1\times10^{-6}\ \text{C} q 3 = + 1 × 1 0 − 6 C at 0.10 m 0.10\ \text{m} 0.10 m right. Answer: F net = 0 N F_{\text{net}}=0\ \text{N} F net = 0 N . Equal charges at equal distances create equal but opposite forces that cancel.
Flashcard 6: Find the force ratio F 2 F 1 \frac{F_2}{F_1} F 1 F 2 if the distance changes from r 1 r_1 r 1 to r 2 = 3 r 1 r_2=3r_1 r 2 = 3 r 1 . Answer: F 2 F 1 = 1 9 \frac{F_2}{F_1}=\frac{1}{9} F 1 F 2 = 9 1 . Since F ∝ 1 / r 2 F \propto 1/r^2 F ∝ 1/ r 2 , tripling r r r reduces force by factor of 9.
Flashcard 7: Find r r r if q 1 = q 2 = + 1 × 10 − 6 C q_1=q_2=+1\times 10^{-6}\ \text{C} q 1 = q 2 = + 1 × 1 0 − 6 C and the force magnitude is F = 1.0 N F=1.0\ \text{N} F = 1.0 N . Answer: r ≈ 0.095 m r\approx 0.095\ \text{m} r ≈ 0.095 m . r = k ∣ q 1 q 2 ∣ F = 8.99 × 10 − 3 1.0 ≈ 0.095 m r = \sqrt{\frac{k|q_1q_2|}{F}} = \sqrt{\frac{8.99\times10^{-3}}{1.0}} \approx 0.095\ \text{m} r = F k ∣ q 1 q 2 ∣ = 1.0 8.99 × 1 0 − 3 ≈ 0.095 m
Flashcard 8: Identify whether the force is attractive or repulsive for q 1 = + 2 μ C q_1=+2\ \mu\text{C} q 1 = + 2 μ C and q 2 = − 5 μ C q_2=-5\ \mu\text{C} q 2 = − 5 μ C . Answer: Attractive. Opposite signs (+ + + and − - − ) mean attractive force.
Flashcard 9: What is the direction of the Coulomb force on each charge relative to the line joining them? Answer: Along the line connecting the charges. Electric forces act along the line between charges.
Flashcard 10: What is the SI value of Coulomb's constant k k k used in Coulomb's law? Answer: k = 8.99 × 10 9 N ⋅ m 2 / C 2 k = 8.99\times 10^9\ \text{N}\cdot\text{m}^2/\text{C}^2 k = 8.99 × 1 0 9 N ⋅ m 2 / C 2 . This constant relates charge, distance, and force in SI units.
Flashcard 11: What is the net force magnitude if two forces of 5 N 5\ \text{N} 5 N and 3 N 3\ \text{N} 3 N act on a charge in opposite directions? Answer: 2 N 2\ \text{N} 2 N . Forces in opposite directions subtract: 5 − 3 = 2 5 - 3 = 2 5 − 3 = 2 N.
Flashcard 12: Identify the vector direction of the force on q 1 q_1 q 1 due to q 2 q_2 q 2 in Coulomb's law problems. Answer: Along the line from q 1 q_1 q 1 to q 2 q_2 q 2 (toward or away). Force acts along the line connecting the charges.
Flashcard 13: What is the net force magnitude if two forces of 3 N 3\ \text{N} 3 N and 5 N 5\ \text{N} 5 N act on a charge in the same direction? Answer: 8 N 8\ \text{N} 8 N . Forces in same direction add: 3 + 5 = 8 3 + 5 = 8 3 + 5 = 8 N.
Flashcard 14: Calculate F F F for q 1 = + 4 × 10 − 6 C q_1=+4\times 10^{-6}\ \text{C} q 1 = + 4 × 1 0 − 6 C , q 2 = + 2 × 10 − 6 C q_2=+2\times 10^{-6}\ \text{C} q 2 = + 2 × 1 0 − 6 C , r = 0.20 m r=0.20\ \text{m} r = 0.20 m . Answer: F ≈ 1.8 N F\approx 1.8\ \text{N} F ≈ 1.8 N . F = 8.99 × 10 9 × 8 × 10 − 12 0.04 ≈ 1.8 N F = 8.99\times10^9 \times \frac{8\times10^{-12}}{0.04} \approx 1.8\ \text{N} F = 8.99 × 1 0 9 × 0.04 8 × 1 0 − 12 ≈ 1.8 N
Flashcard 15: What is the sign of the force between like charges (both + + + or both − - − )? Answer: Repulsive. Like charges repel each other.
Flashcard 16: What happens to F F F if both charges change from ( q 1 , q 2 ) (q_1,q_2) ( q 1 , q 2 ) to ( 2 q 1 , 3 q 2 ) (2q_1,3q_2) ( 2 q 1 , 3 q 2 ) with r r r unchanged? Answer: F → 6 F F\to 6F F → 6 F . Force scales with product of charges: 2 × 3 = 6 2 \times 3 = 6 2 × 3 = 6 .
Flashcard 17: State the formula for the magnitude of the electric force between two point charges. Answer: F = k ∣ q 1 q 2 ∣ r 2 F = k\frac{|q_1 q_2|}{r^2} F = k r 2 ∣ q 1 q 2 ∣ . Force is proportional to charge product and inversely proportional to distance squared.
Flashcard 18: What is the direction of the force between two charges with opposite signs? Answer: Attractive; along the line joining the charges. Opposite charges attract each other directly along their connecting line.
Flashcard 19: What happens to F F F if the separation distance changes from r r r to 2 r 2r 2 r (charges unchanged)? Answer: F → F 4 F\to \frac{F}{4} F → 4 F . Doubling distance reduces force by factor of 2 2 = 4 2^2 = 4 2 2 = 4 .
Flashcard 20: What is the value of Coulomb's constant k k k in N ⋅ m 2 / C 2 \text{N}\cdot\text{m}^2/\text{C}^2 N ⋅ m 2 / C 2 ? Answer: k = 8.99 × 10 9 N ⋅ m 2 / C 2 k = 8.99\times 10^9\ \text{N}\cdot\text{m}^2/\text{C}^2 k = 8.99 × 1 0 9 N ⋅ m 2 / C 2 . This fundamental constant relates charge, distance, and force in SI units.
Flashcard 21: For three collinear charges, what principle tells you the net force on one charge is the vector sum of pairwise forces? Answer: Superposition: F ⃗ net = ∑ F ⃗ i \vec{F}_{\text{net}}=\sum \vec{F}_i F net = ∑ F i . Vector sum of individual forces gives the net force on a charge.
Flashcard 22: What is the net force magnitude on q 3 q_3 q 3 if two equal forces of 5 N 5\ \text{N} 5 N act on it in opposite directions? Answer: 0 N 0\ \text{N} 0 N . Equal forces in opposite directions cancel out.
Flashcard 23: State Newton's third law relationship for the forces two charges exert on each other. Answer: F ⃗ 12 = − F ⃗ 21 \vec{F}_{12}=-\vec{F}_{21} F 12 = − F 21 . Forces form an action-reaction pair with equal magnitude, opposite direction.
Flashcard 24: What happens to F F F if the separation distance changes from r r r to 2 r 2r 2 r ? Answer: F → F 4 F\rightarrow\frac{F}{4} F → 4 F . Force decreases by factor of r 2 r^2 r 2 , so doubling r r r gives F / 4 F/4 F /4 .
Flashcard 25: Identify the SI unit of electric charge used in Coulomb's law calculations. Answer: coulomb (C) \text{coulomb (C)} coulomb (C) . The standard unit for measuring electric charge in the SI system.
Flashcard 26: What is the sign of the force between opposite charges (one + + + and one − - − )? Answer: Attractive. Opposite charges attract each other.
Flashcard 27: State the formula for the magnitude of the electrostatic force between two point charges. Answer: F = k ∣ q 1 q 2 ∣ r 2 F = k\frac{|q_1 q_2|}{r^2} F = k r 2 ∣ q 1 q 2 ∣ . Force is proportional to charge product and inversely proportional to distance squared.
Flashcard 28: Identify the SI unit of electric charge used in Coulomb's law. Answer: coulomb (C) \text{coulomb (C)} coulomb (C) . Named after Charles-Augustin de Coulomb who discovered the law.
Flashcard 29: What happens to F F F if the separation distance changes from r r r to r 3 \frac{r}{3} 3 r (charges unchanged)? Answer: F → 9 F F\to 9F F → 9 F . Reducing distance by 3 increases force by factor of 3 2 = 9 3^2 = 9 3 2 = 9 .
Flashcard 30: Find the force ratio F 2 F 1 \frac{F_2}{F_1} F 1 F 2 if q 1 q_1 q 1 is tripled and q 2 q_2 q 2 is halved while r r r stays constant. Answer: F 2 F 1 = 3 2 \frac{F_2}{F_1}=\frac{3}{2} F 1 F 2 = 2 3 . Force ratio equals charge product ratio: ( 3 ) ( 1 / 2 ) = 3 / 2 (3)(1/2) = 3/2 ( 3 ) ( 1/2 ) = 3/2 .
Flashcard 31: What happens to F F F if the separation distance changes from r r r to r 3 \frac{r}{3} 3 r ? Answer: F → 9 F F\rightarrow 9F F → 9 F . Force increases by factor of r 2 r^2 r 2 , so r / 3 r/3 r /3 gives 9 F 9F 9 F .
Flashcard 32: Identify the SI unit of electric force in Coulomb's law problems. Answer: newton (N) \text{newton (N)} newton (N) . Force is measured in newtons in the SI system.
Flashcard 33: What happens to F F F if q 1 q_1 q 1 is doubled while q 2 q_2 q 2 and r r r stay the same? Answer: F → 2 F F\rightarrow 2F F → 2 F . Force is directly proportional to each charge.
Flashcard 34: State Newton's third-law relationship for the forces between two point charges. Answer: F ⃗ 12 = − F ⃗ 21 \vec{F}_{12} = -\vec{F}_{21} F 12 = − F 21 . Forces form an action-reaction pair with equal magnitude, opposite direction.