Q1: What is the electric force of repulsion between two electrons at a distance of 1 m?
Given:
Charge of electron, q₁ = q₂ = 1.6 × 10⁻¹⁹ C
Distance, r = 1 m
Coulomb's constant, k = 9 × 10⁹ Nm²/C²
Formula: F = k × (q₁ × q₂) / r²
Calculation:
F = (9 × 10⁹) × (1.6 × 10⁻¹⁹ × 1.6 × 10⁻¹⁹) / (1)²
F = (9 × 10⁹) × (2.56 × 10⁻³⁸)
F = 23.04 × 10⁻²⁹
F = 2.3 × 10⁻²⁸ N
Answer: 2.3 × 10⁻²⁸ N
Charge of electron, q₁ = q₂ = 1.6 × 10⁻¹⁹ C
Distance, r = 1 m
Coulomb's constant, k = 9 × 10⁹ Nm²/C²
Formula: F = k × (q₁ × q₂) / r²
Calculation:
F = (9 × 10⁹) × (1.6 × 10⁻¹⁹ × 1.6 × 10⁻¹⁹) / (1)²
F = (9 × 10⁹) × (2.56 × 10⁻³⁸)
F = 23.04 × 10⁻²⁹
F = 2.3 × 10⁻²⁸ N
Answer: 2.3 × 10⁻²⁸ N
Q2: Two point charges q₁ = 5μC and q₂ = 3μC are placed at a distance of 5 cm. What will be the Coulomb's force between them?
Given:
q₁ = 5 μC = 5 × 10⁻⁶ C
q₂ = 3 μC = 3 × 10⁻⁶ C
Distance, r = 5 cm = 0.05 m
k = 9 × 10⁹ Nm²/C²
Formula: F = k × (q₁ × q₂) / r²
Calculation:
F = (9 × 10⁹) × (5 × 10⁻⁶ × 3 × 10⁻⁶) / (0.05)²
F = (9 × 10⁹) × (15 × 10⁻¹²) / (0.0025)
F = (135 × 10⁻³) / 0.0025
F = 0.135 / 0.0025
F = 54 N
Answer: 54 N
q₁ = 5 μC = 5 × 10⁻⁶ C
q₂ = 3 μC = 3 × 10⁻⁶ C
Distance, r = 5 cm = 0.05 m
k = 9 × 10⁹ Nm²/C²
Formula: F = k × (q₁ × q₂) / r²
Calculation:
F = (9 × 10⁹) × (5 × 10⁻⁶ × 3 × 10⁻⁶) / (0.05)²
F = (9 × 10⁹) × (15 × 10⁻¹²) / (0.0025)
F = (135 × 10⁻³) / 0.0025
F = 0.135 / 0.0025
F = 54 N
Answer: 54 N
Q3: If 2 μC charge is placed in the field of 3.42 × 10¹¹ N/C, what will be the force on it?
Given:
Charge, q = 2 μC = 2 × 10⁻⁶ C
Electric field, E = 3.42 × 10¹¹ N/C
Formula: F = q × E
Calculation:
F = (2 × 10⁻⁶) × (3.42 × 10¹¹)
F = 6.84 × 10⁵
F = 684 × 10³ N
F = 6.84 × 10⁵ N
Answer: 6.84 × 10⁵ N
Charge, q = 2 μC = 2 × 10⁻⁶ C
Electric field, E = 3.42 × 10¹¹ N/C
Formula: F = q × E
Calculation:
F = (2 × 10⁻⁶) × (3.42 × 10¹¹)
F = 6.84 × 10⁵
F = 684 × 10³ N
F = 6.84 × 10⁵ N
Answer: 6.84 × 10⁵ N
Q4: What is the charge on the capacitor, if a 40 μF capacitor has a potential difference of 6 V across it?
Given:
Capacitance, C = 40 μF = 40 × 10⁻⁶ F
Potential difference, V = 6 V
Formula: Q = C × V
Calculation:
Q = (40 × 10⁻⁶) × 6
Q = 240 × 10⁻⁶
Q = 2.4 × 10⁻⁴ C
Answer: 2.4 × 10⁻⁴ C
Capacitance, C = 40 μF = 40 × 10⁻⁶ F
Potential difference, V = 6 V
Formula: Q = C × V
Calculation:
Q = (40 × 10⁻⁶) × 6
Q = 240 × 10⁻⁶
Q = 2.4 × 10⁻⁴ C
Answer: 2.4 × 10⁻⁴ C
Q5: The potential difference between two points is 100 V. If an unknown charge is moved between these points, the amount of work done is 500 J. Find the amount of charge.
Given:
Potential difference, V = 100 V
Work done, W = 500 J
Formula: V = W / q
q = W / V
Calculation:
q = 500 / 100
q = 5 C
Answer: 5 C
Potential difference, V = 100 V
Work done, W = 500 J
Formula: V = W / q
q = W / V
Calculation:
q = 500 / 100
q = 5 C
Answer: 5 C
Q6: Find the equivalent capacitance when a 4μF, 3μF and 2μF capacitor are connected in series.
Given:
C₁ = 4 μF
C₂ = 3 μF
C₃ = 2 μF
Connected in series
Formula for series combination:
1/Cₑq = 1/C₁ + 1/C₂ + 1/C₃
Calculation:
1/Cₑq = 1/4 + 1/3 + 1/2
1/Cₑq = 0.25 + 0.333 + 0.5
1/Cₑq = 1.083
Cₑq = 1/1.083
Cₑq = 0.92 μF
Answer: 0.92 μF
C₁ = 4 μF
C₂ = 3 μF
C₃ = 2 μF
Connected in series
Formula for series combination:
1/Cₑq = 1/C₁ + 1/C₂ + 1/C₃
Calculation:
1/Cₑq = 1/4 + 1/3 + 1/2
1/Cₑq = 0.25 + 0.333 + 0.5
1/Cₑq = 1.083
Cₑq = 1/1.083
Cₑq = 0.92 μF
Answer: 0.92 μF
