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Question

Exercise:
In a parallel plate capacitor with air between the plates, each plate has an area of 6×103m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?

Explain what would happen if in the capacitor given in Exercise a 3mm 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
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C=ϵoAd
According to the values given C=1.8×1011F
Now Q=CV
Hence charge on the positive plate will be 1.8×109C
(a) Dielectric constant, k = 6
C=1.8×1011 F
new capacitance, C=kC=108 pF
V=100volts
q=V×C=1.08×108 C
V remains the same.

(b) C=kC=6×1.8×1011=108pF
q remains the same.
q=1.8×109 C
V=q/C=16.67 V


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Explain what would happen if in the capacitor given in Exercise a 3mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,
(a) While the voltage supply remained connected.
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