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Standard XII
Physics
Capacitance of Parallel Plate Capacitor
Question

A parallel plate capacitor has a capacity 80×10−6 F when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant 20. The capacitor is now connected to a battery of 30 V by wires. The dielectric slab is then removed. Then, the charge that passes now through the wire is :
  1. 45.6×10−3 C
  2. 120×10−3 C
  3. 25.3×10−3 C
  4. 12×10−3 C

A
45.6×10−3 C
B
12×10−3 C
C
120×10−3 C
D
25.3×10−3 C
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Solution
Verified by Toppr

Charge passing through the wire, Δq=ΔCV
=(C′−C)V=(k−1)CV=(20−1)(80×10−6)(30)=45.6×10−3
So, charge passing through the wire =45.6×10−3 C

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