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.
Solution
(a) Dielectric constant of the mica sheet, k = 6
Initial capacitance, C = 1.771 × 10−11 F
New Capacitance, C''=kC=`6xx1.771xx10^-11=106 pF`
Supply voltage, V = 100 V
New Capacitance, q'=C'V=`6xx1.771xx10^-9=1.06 xx 10^-8pF`
Potential across the plates remains 100 V.
(b) Dielectric constant, k = 6
Initial capacitance, C = 1.771 × 10−11 F
New Capacitance, C''=kC=`6xx1.771xx10^-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,
`therefore V'=q/C'`
`=(1.771xx10^-9)/(106xx10^-12)`
=16.7 V
