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Asked by Prasenjit Last Modified
Answer 
Apoorva Purohit
Area of each plate of the parallel plate capacitor, A = 6 × 10−3 m2
Distance between the plates, d = 3 mm = 3 × 10−3 m
Supply voltage, V = 100 V
Capacitance C of a parallel plate capacitor is given by,
Where,
= Permittivity of free space
= 8.854 × 10−12 N−1 m−2 C−2
Therefore, capacitance of the capacitor is 17.71 pF and charge on each plate is 1.771 × 10−9 C.
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Shivank Sharma
Engineer and a civil services exam aspirant.
The area of the parallel plate capacitor is , A=6×10−3m2
Distance between the plates, d=3mm=3×10−3m
Supply voltage, V = 100 V
Capacitance C , C=E0Ad
where ∈0= Permittivity of free space =8.854×10−12 F/m
C=(8.854×10^−12) × (6×10^−3)× (3×10−3) =17.71×10^−12F
So, charge on each plate of the capacitor q=VC=100×17.71×10 ^−12
Hence, capacitance of the capacitor is 17.71 pF and charge on each plate is 1.771×10−9C.
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