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

Answer 1

Answer:

Explanation:

[ You have to insert the previous exercise question 2.8 here for my reference]

(1) The capacitance of the capacitor will increase when the mica sheet is inserted, so the charge on the capacitor also increases.

(2) when the voltage is disconnected, then the charge on the capacitor remains constant. But due to an increase in capacitance,  the potential difference across the capacitor will decrease( as, V = Q/C).

Detailed Solution:

We know, capacitance = dielectric constant x $C_{0}$

$C = K C_{0}$   where K is the dielectric constant of the mica sheet.

C = 6 x 18 pF = 108 pF

(1) 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 x 100 V

$Q = 108 \times 10^{-12} \times 100 C = 1.08 \times 10^{-8} C$

(2) If the voltage supply was disconnected, then the charge on the capacitor remains the same. e.g., $Q = 1.8 \times 10^{-9} C$    

And hence the potential difference across the capacitor becomes

$V = Q/C = 1.8 \times 10^{-9} / 1.08 \times 10^{-10} = 16.6 V$

Answer 2

Explanation:

We know, capacitance = dielectric constant x C_{0}C

0

C = K C_{0}C=KC

0

where K is the dielectric constant of the mica sheet.

C = 6 x 18 pF = 108 pF

(1) 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 x 100 V