Q. 9 4.2( 13 Votes )

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) 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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