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Question
Explain what would happen if the capacitor given in previous question a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,
- While the voltage supply remained connected.
- After the supply was disconnected.
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Solution
(a) Dielectric constant of the mica sheet, k = 6
Initial capacitance, C = 1.77 × 10−11 F
New Capacitance, C' = kC
= 6 × 1.77 × 10−11
= 106 pF
Supply voltage, V = 100 V
New Capacitance, q' = C'V
= 106 × 100 pC
= 1.06 × 10−8 C
Potential across the plates remains 100 V.
(b) Dielectric constant, k = 6
Initial capacitance, C = 1.77 × 10−11 F
New Capacitance, C' = kC
= 6 × 1.77 × 10−11
= 106 pF
If the supply voltage is removed, then there will be no effect on the amount of charge in the plates.
Charge = 1.77 × 10−9 C
Potential across the plates comes from,
∴ `V' = q/(C')`
= `(1.77 xx 10^-9)/(106 xx 10^-12)`
= 16.7 V
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