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An air-filled parallel-plate capacitor has plates of area 2.50 cm² separated by 2.00 mm. The capacitor is connected to a 3.0-V battery. (a) Find the value of its capacitance in pF

User Teefour
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Final answer:

The capacitance of the given air-filled parallel-plate capacitor, with an area of 2.50 cm² and a separation of 2.00 mm when connected to a 3.0-V battery, is calculated to be 1110 picoFarads (pF) using the formula for the capacitance of a parallel-plate capacitor.

Step-by-step explanation:

The subject of the question is the calculation of the capacitance of a parallel-plate capacitor in the physical context. The formula for the capacitance C of a parallel-plate capacitor is given by C = ε_0 × (A/d), where ε_0 is the vacuum permittivity (ε_0 = 8.854 x 10^{-12} F/m), A is the area of the plates in square meters, and d is the separation between the plates in meters.

For the given air-filled parallel-plate capacitor:

  • Area A = 2.50 cm² = 2.50 x 10^{-4} m²,
  • Separation d = 2.00 mm = 2.00 x 10^{-3} m.

Substituting these into the capacitance formula gives:

C = 8.854 x 10^{-12} F/m × (2.50 x 10^{-4} m² / 2.00 x 10^{-3} m)
C = 1.11 x 10^{-12} F
C = 1110 pF

Therefore, the capacitance of the capacitor is 1110 pF (picoFarads).

User Typedfern
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