Question

Explain what would happen if the capacitor given in previous question a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,

1. While the voltage supply remained connected.
2. After the supply was disconnected.

Answer 1

1. Dielectric constant of the mica sheet, k = 6
   Initial capacitance, C = 1.771 × 10⁻¹¹ F
   New Capacitance, C' = kC
   = 6 × 1.771 × 10⁻¹¹
   = 106 pF
   Supply voltage, V = 100 V
   New Capacitance, q' = C'V
   = 6 × 1.771 × 10⁻⁹
   = 1.06 × 10⁻⁸ C
   Potential across the plates remains 100 V.
2. Dielectric constant, k = 6
   Initial capacitance, C = 1.771 × 10⁻¹¹ F
   New Capacitance, C' = kC
   = 6 × 1.771 × 10⁻¹¹
   = 106 pF
   If the supply voltage is removed, then there will be no effect on the amount of charge in the plates.
   Charge = 1.771 × 10⁻⁹ C
   Potential across the plates comes from,
   ∴ `"V'" = "q"/"C'"`
   = `(1.771 xx 10^-9)/(106 xx 10^-12)`
   = 16.7 V