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.

(a) When the voltage supply remains connected:

The capacitance of the capacitor will become K times.


Therefore, C’ = kC


Where k = dielectric constant = 6 × 17.7pF = 106.2 pF


The potential difference across the two plates of the capacitor will remain equal to the supply voltage i.e. 100 V


The charge on the capacitor,


q’ = C’V = 160.2 x 10-12 x 100


= 1.602 x 10-8 C


(b) After the voltage supply is disconnected:


As calculated above, the capacitance of the capacitor, C’ = 106.2 pF


The potential difference will decrease on introducing mica sheet by a factor of K,


⇒


The charge on the capacitor,


q’ = C’V’ = 106.2 × 10-12 × 16.67


q’ = 1.77 x 10-9 C


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