Answer :
The capacitance of two square, parallel conducting plates with an area of 5.00 cm by 5.00 cm and separated by 0.100 mm is 221 pF (picofarads). This is calculated using the capacitance formula for a parallel-plate capacitor.
The student is asking to find the capacitance of a parallel-plate capacitor with specific dimensions. For two square, parallel conducting plates with each having an area of 5.00 cm x 5.00 cm and separated by 0.100 mm, we utilize the capacitance formula for a parallel-plate capacitor:
C = rac{ ext{ε}_0 imes A}{d}
Where C is the capacitance, ε₀ is the permittivity of free space (8.85 x 10-12 C2/N·m2), A is the area of one of the plates, and d is the separation distance between the plates.
Firstly, we need to convert the area from cm2 to m2 by multiplying by (10-2)2, so the area A = (5.00 cm x 10-2 m/cm) x (5.00 cm x 10-2 m/cm) = 0.0025 m2. The separation distance d must also be converted from mm to m by multiplying by 10-3, so d = 0.100 mm x 10-3 m/mm = 0.0001 m.
Now we can calculate the capacitance:
C = rac{8.85 x 10-12 C2/N·m2 x 0.0025 m2}{0.0001 m}
C = 2.21 x 10-13 F or 221 pF
The capacitance of the parallel-plate capacitor is therefore 221 pF (picofarads).
