Coulomb Law Answer Key Document
Electrostatics // Answer Key & Teaching Guide
REF: WS_COULOMB_02-A_KEY
Coulomb Law Workshop
Teacher Resource Official Solutions
Key Equation
\( F_e = k \frac{|q_1 q_2|}{r^2} \)
Constants
\( k = 9 \times 10^9 \)
\( e = 1.6 \times 10^{-19} \, \text{C} \)
Conversions
\( 1 \, \mu\text{C} = 10^{-6} \, \text{C} \)
\( 1 \, \text{m} = 100 \, \text{cm} \)
1
Magnitude Calculations
Problem 1: Two point charges (\(+2.5 \, \mu\text{C}\) and \(-4.0 \, \mu\text{C}\)) at \(0.5 \, \text{m}\).
Step 1: Variables
\( q_1 = 2.5 \times 10^{-6} \, \text{C} \), \( q_2 = 4.0 \times 10^{-6} \, \text{C} \), \( r = 0.5 \, \text{m} \)
Step 2: Math
\[ F_e = \frac{(9 \times 10^9)(2.5 \times 10^{-6})(4.0 \times 10^{-6})}{(0.5)^2} = \frac{0.09}{0.25} = 0.36 \, \text{N} \]
Key: 0.36 N | Attractive
Common Error: Forgetting to square 0.5. Divide by 0.25, not 0.5.
Problem 2: Alpha particle (\(+2e\)) and electron (\(-e\)) at \(2.0 \times 10^{-10} \, \text{m}\).
Step 1: Charges
\( q_1 = 2(1.6 \times 10^{-19}) = 3.2 \times 10^{-19} \, \text{C} \); \( q_2 = 1.6 \times 10^{-19} \, \text{C} \)
Step 2: Math
\[ F_e = \frac{(9 \times 10^9)(3.2 \times 10^{-19})(1.6 \times 10^{-19})}{(2.0 \times 10^{-10})^2} = \frac{46.08 \times 10^{-29}}{4.0 \times 10^{-20}} \]
\[ F_e = 11.52 \times 10^{-9} = 1.15 \times 10^{-8} \, \text{N} \]
Key: \( 1.15 \times 10^{-8} \, \text{N} \) | Attractive
COULOMB LAW WORKSHOP PAGE 1 OF 2
Analysis & Vector Solutions
Part 2 of 2
2
The Inverse-Square Rule
Scenario: Distance x4
If \( r \to 4r \), then \( F \to 1/4^2 \).
Answer: Force decreases by factor of 16.
Scenario: Charges x2, Dist x2
\( F \propto \frac{(2)(2)}{(2)^2} = \frac{4}{4} = 1 \).
Answer: Force remains unchanged.
3
Vector Command
Part A: Force from Q1 on Q2
\( r_{12} = 2.0 \, \text{m} \)
\[ F_{12} = \frac{(9 \times 10^9)(5 \times 10^{-6})(2 \times 10^{-6})}{2^2} = 0.0225 \, \text{N} \]
Direction: Attractive. \( q_2 \) is pulled toward \( q_1 \) (LEFT / -x).
Part B: Force from Q3 on Q2
\( r_{32} = 3.0 \, \text{m} \)
\[ F_{32} = \frac{(9 \times 10^9)(3 \times 10^{-6})(2 \times 10^{-6})}{3^2} = 0.0060 \, \text{N} \]
Direction: Attractive. \( q_2 \) is pulled toward \( q_3 \) (RIGHT / +x).
Part C: Vector Sum (\( F_{net} \))
\( F_{net} = +0.0060 - 0.0225 = -0.0165 \, \text{N} \)
Final Result: 0.0165 N to the Left
Vector Tip: Signs in Coulomb's Law math are often confusing. Best practice: Use magnitudes for math, then assign directions based on attraction/repulsion logic.
COULOMB LAW WORKSHOP PAGE 2 OF 2
