In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10−3 m2 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?

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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...
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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. read less
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Capacitance of parallel plate capacitor C= €A/d and charge on the capacitor Q=CV. Put the values and get the answer.
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Engineer and a civil services exam aspirant.

Area of each plate of the parallel plate capacitor, A=6×10−3m2 Distance between the plates, d=3mm=3×10−3md=3mm=3×10−3m Supply voltage, V = 100 V Capacitance C of a parallel plate capacitor is given by, C=E0Ad where ∈0∈0= Permittivity of free...
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Area of each plate of the parallel plate capacitor,A=6×10−3m2 Distance between the plates,d=3mm=3×10−3md=3mm=3×10−3m Supply voltage, V = 100 V Capacitance C of a parallel plate capacitor is given by,C=E0Ad where ∈0∈0= Permittivity of free space=8.854×10−12N^−1m^−2C^−2 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 Therefore, capacitance of the capacitor is 17.71 pF and charge on each plate is1.771×10−9C. read less
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Mechanical engineer with 2 years of teaching experience

The area of the parallel plate capacitor is , A=6×10−3m2 Distance between the plates, d=3mm=3×10−3mSupply voltage, V = 100 V Capacitance C , C=E0Adwhere ∈0= Permittivity of free space =8.854×10−12 F/m C=(8.854×10^−12) × (6×10^−3)×...
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The area of the parallel plate capacitor is , A=6×10−3m2 Distance between the plates, d=3mm=3×10−3mSupply voltage, V = 100 V Capacitance C , C=E0Adwhere ∈0= Permittivity of free space =8.854×10−12 F/mC=(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. read less
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17.71 pF , Charge on each plate in 1.771 X 10-9 C
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NEET UG PHYSICS, IIT JEE MAINS and also any board level

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17.71pF
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