Question 2.9: Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates, (a) While the voltage supply remained connected. (b) After the supply was disconnected.
Class 12 - Physics - Electrostatic Potential And Capacitance Electrostatic Potential And Capacitance Page-88
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We know, capacitance = dielectric constant × C₀
C = KC₀ = 6 × 18pF = 108pF [ see question - 2.8]
(a) if the voltage supply remained connected , then the potential difference across the capacitor will remain the same. e.g., V = 100V and hence, charge on the capacitor becomes Q = CV = 108pF × 100V
Q = 108 × 10⁻¹² × 100 = 1.08 × 10⁻⁸ C
(b) if the voltage supply was disconnected , then charge on the capacitor reamins the same. e.g., Q = 1.8 × 10⁻⁹C [ see question - 2.8]
And hence the potential difference across the capacitor becomes
V = Q/C = 1.8 × 10⁻⁹/1.08 × 10⁻¹⁰ = 16.6V
C = KC₀ = 6 × 18pF = 108pF [ see question - 2.8]
(a) if the voltage supply remained connected , then the potential difference across the capacitor will remain the same. e.g., V = 100V and hence, charge on the capacitor becomes Q = CV = 108pF × 100V
Q = 108 × 10⁻¹² × 100 = 1.08 × 10⁻⁸ C
(b) if the voltage supply was disconnected , then charge on the capacitor reamins the same. e.g., Q = 1.8 × 10⁻⁹C [ see question - 2.8]
And hence the potential difference across the capacitor becomes
V = Q/C = 1.8 × 10⁻⁹/1.08 × 10⁻¹⁰ = 16.6V
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