In a parallel plate capacitor with air between the plates, each plate has an area of 6×10−3m2 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.
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Solution
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C=ϵoAd
According to the values given C=1.8×10−11F
Now Q=CV
Hence charge on the positive plate will be 1.8×10−9C
(a) Dielectric constant, k = 6
C=1.8×10−11F new capacitance, C′=kC=108pF V=100volts q′=V×C′=1.08×10−8C V remains the same.
(b) C′=kC=6×1.8×10−11=108pF
q remains the same. q=1.8×10−9C V′=q/C′=16.67V
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Q1
Exercise:
In a parallel plate capacitor with air between the plates, each plate has an area of 6×10−3m2 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.
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Q2
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,