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Class 12 › Physics › Electric Potential & Capacitance
CBSE2021Class Class 12 · Physics3 Marks · Short✅ Verified
Calculate the voltage needed to balance an oil drop carrying 10 electrons, between capacitor plates 5 mm apart. Mass = 3×10⁻¹⁶ kg, e = 1.6×10⁻¹⁹ C, g = 9.8 m/s².
✅ Answer & Solution
Given:
• Number of electrons, n = 10
• Charge on drop, q = ne = 10 × 1.6 × 10⁻¹⁹ = <b>1.6 × 10⁻¹⁸ C</b>
• Plate separation, d = 5 mm = 5 × 10⁻³ m
• Mass of drop, m = 3 × 10⁻¹⁶ kg
• g = 9.8 m/s²
Condition for Balance:
Electric Force = Gravitational Force
qE = mg
q(V/d) = mg
Solving for V:
V = mgd / q
V = (3 × 10⁻¹⁶ × 9.8 × 5 × 10⁻³) / (1.6 × 10⁻¹⁸)
V = (147 × 10⁻¹⁹) / (1.6 × 10⁻¹⁸)
V = 147 / 16
V ≈ <b>9.19 V ≈ 9.2 V</b>
