What is the capacitance of a parallel plate capacitor, given the plates measure 6 cm by 7 cm and are separated by a distance of 0.003 cm?
Final answer: The capacitance of a parallel plate capacitor with 6 cm by 7 cm plates and a separation of 0.003 cm is approximately 12.4 pF, rounded to the nearest available option is (a) 12 pF. Option A is correct.
Calculation of Capacitance:
Capacitance Formula: The formula for the capacitance of a parallel plate capacitor is C = ε₀ * (A/d), where C is the capacitance, ε₀ is the permittivity of free space, A is the area of one of the plates, and d is the separation between the plates.
Given Data:
- Plate dimensions: 6 cm by 7 cm
- Plate separation: 0.003 cm
Calculation Steps:
- Convert the area and separation into meters:
- Area A = 6 cm x 7 cm = 0.06 m x 0.07 m = 0.0042 m²
- Separation d = 0.003 cm = 0.00003 m
- Calculate the capacitance:
- C = ε₀ * (A/d)
- C = 8.854 x 10^-12 F/m * (0.0042 m² / 0.00003 m)
- C = 12.4 x 10^-12 F, or 12.4 pF
Therefore, the capacitance of the parallel plate capacitor given the specified plate dimensions and separation distance is approximately 12.4 pF, which rounds to the closest available option of 12 pF. This makes option A the correct answer.