Answer :
Answer:
a) the capacitance:
The capacitance of a parallel plate capacitor can be calculated by using the formula,
[tex] \sf \: C = \dfrac{\varepsilon_o A}{{d}} [/tex]
where,
- [tex] \varepsilon_o[/tex] is the permittivity of the material between the plates,
- A is the area of the plates, and
- d is the separation between the plates.
We are given,
El-plate air-filled capacitor having area 40 cm² and plate spacing 1.0 mm is charged to a potential difference of 600 V.
- Area of the plates, A = 40 cm² = [tex] \sf 40 \times 10^{-4} m^2[/tex]
- Permittivity, [tex] \sf \varepsilon_o = 8.85 \times 10^{-12} C^2/N m^2 [/tex]
- separation between the plates, d = 1.0 mm = [tex] \sf 1.0 \times 10^{-3} \: m [/tex]
Substituting the required values,
[tex] \sf \: C = \dfrac{ 8.85 \times 10^{-12} C^2/N \: m^2 \times 40 \times 10^{-4} m^2}{1.0 \times 10^{-3} \: m} \\ [/tex]
[tex]\sf C = 3.5 \times 10^{-11 } \ F [/tex]
b) The charge on each plate:
[tex] \sf q = CV [/tex]
where,
- C is the capacitance of the capacitor, and
- V is the voltage across the capacitor.
substituting the required values,
[tex] \sf q = (3.5 \times 10^{-11 } \ F) ( 600 V) \\ [/tex]
[tex] \sf q = 2.5 \times 10^{-8} \ C[/tex]
Therefore,
a) The capacitance of the parallel plate capacitor is [tex]\sf 3.5 \times 10^{-11 } \ F[/tex]
b) The required charge on each plate is [tex] \sf 2.5 \times 10^{-8} \ C[/tex]
The capacitance of the given parallel-plate air-filled capacitor is 35.4 pF and the charge on each plate is 21.2 nC.
Capacitance and Charge of a Parallel-Plate Capacitor
To calculate the capacitance (C) of a parallel-plate capacitor, we use the formula:
C = rac{ arepsilon_0 A}{d}
where arepsilon_0 is the vacuum permittivity ( arepsilon_0 = 8.85 imes 10^{-12} F/m), A is the area of the plates in square meters, and d is the distance between the plates in meters.
For part (a), converting the area to square meters (40 cm squared = 4 imes 10^{-3} m squared) and the distance to meters (1.0 mm = 1 imes 10^{-3} m), we find the capacitance using the formula:
C = rac{8.85 imes 10^{-12} imes 4 imes 10^{-3}}{1 imes 10^{-3}}
C = 3.54 imes 10^{-11} F or 35.4 pF.
For part (b), the charge (Q) on each plate can be calculated using the formula:
Q = C imes V
where V is the potential difference. Substituting in the values:
Q = 3.54 imes 10^{-11} F imes 600 V
Q = 2.12 imes 10^{-8} C or 21.2 nC.
