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Difficulty-Based Practice Sets
Level 1: Foundation (Target: 90% accuracy, 1 min/question)
Practice Set 1: Coulomb's Law Basics
Q1. Two charges +4 μC and -2 μC are 20 cm apart. Calculate force between them.
Show Solution
Given: q₁ = 4 × 10⁻⁶ C, q₂ = -2 × 10⁻⁶ C, r = 0.2 m
Formula: F = k|q₁q₂|/r²
F = (9×10⁹)(4×10⁻⁶)(2×10⁻⁶)/(0.2)²
F = 72×10⁻³/0.04 = 1.8 N (attractive)
Q2. Electric field at 10 cm from a +5 nC charge is:
(a) 450 N/C (b) 900 N/C (c) 4500 N/C (d) 9000 N/C
Show Solution
E = kQ/r² = (9×10⁹)(5×10⁻⁹)/(0.1)²
E = 45/0.01 = 4500 N/C
Answer: (c)
Q3. Two identical spheres with charges +6 μC and -2 μC touch and separate. Final charge on each?
Show Solution
Total charge = +6 + (-2) = +4 μC
After contact, equally distributed
Each gets: +4/2 = +2 μC
Q4. Charge on electron is 1.6×10⁻¹⁹ C. How many electrons make -1 μC?
Show Solution
n = Q/e = (1×10⁻⁶)/(1.6×10⁻¹⁹)
n = 6.25 × 10¹² electrons
Q5. Electric potential at 5 cm from +2 nC charge:
Show Solution
V = kQ/r = (9×10⁹)(2×10⁻⁹)/0.05
V = 18/0.05 = 360 V
Level 2: JEE Main (Target: 75% accuracy, 2-3 min/question)
Practice Set 2: Superposition & Vector Addition
Q1. Three charges +q, +q, -2q are at vertices of equilateral triangle (side a). Net force on -2q is:
(a) kq²/a² (b) 2kq²/a² (c) √3 kq²/a² (d) 3kq²/a²
Show Solution
Step 1: Force from each +q on -2q = k(q)(2q)/a² = 2kq²/a²
Step 2: Two forces at 60° angle
Step 3: Resultant = 2F cos(30°) = 2(2kq²/a²)(√3/2)
Answer: (c) √3 kq²/a²
Q2. Electric dipole of moment p is in uniform field E. Maximum torque experienced is:
(a) pE (b) pE/2 (c) 2pE (d) Zero
Show Solution
τ = pE sin θ
Maximum when sin θ = 1 (θ = 90°)
Answer: (a) pE
Q3. Work done to move +2 μC charge from point A (V = 100 V) to point B (V = 200 V):
Show Solution
W = q(V_B - V_A)
W = (2×10⁻⁶)(200 - 100)
W = 2×10⁻⁴ J = 0.2 mJ
Level 3: JEE Advanced (Target: 50-60% accuracy, 5-8 min/question)
Practice Set 3: Advanced Multi-Concept
Q1. A uniformly charged ring of radius R has total charge Q. An electron is released from rest at distance x from center on axis. Find speed when it reaches center. (Given: x >> R)
Show Hint
Hint 1: Use energy conservation
Hint 2: Potential at axis: V = kQ/√(x²+R²)
Hint 3: At center: V = kQ/R
Hint 4: For x >> R, initial V ≈ kQ/x
Q2. Two point charges +Q and -Q are fixed at distance 2a apart. A third charge +q (mass m) is placed at midpoint and displaced slightly perpendicular to line joining the charges. Prove it executes SHM and find time period.
Show Approach
Step 1: Let displacement = y (perpendicular)
Step 2: Calculate forces from both charges
Step 3: Net force will be restoring (∝ -y) for small y
Step 4: F = -ky where k = 2kQq/a³
Step 5: T = 2π√(m/k) = 2π√(ma³/2kQq)
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