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Question:

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.

Answer:

(a) Dielectric constant of the mica sheet, k = 6

Initial capacitance, C = 1.771 × 10 -11 F

New Capacitance, C'= kC= 6x1.771x10-11 =106 pF

Supply voltage, V = 100 V

New Charge, q'=C'V=6x1.771x10-9 =1.06x 10-8 C

Potential across the plates remains 100 V

(b) Dielectric constant, k = 6

Initial capacitance, C = 1.771 × 10-11 F

New Capacitance, C'= kC= 6x1.771x10-11 =106 pF

If 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 is given by,


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