A certain parallel plate capacitor consists of two identical aluminium plates each of area 2times 10^-4 m^2.the plates are separated by a distance of 0,03mm, with air occupying the space between the plates. 1-CALCULATE THE CAPACITANCE OF THE CAPACITOR. AND 2-CALCULATE THE CHARGE STORED ON THE PLATES OF THE CAPACITOR WHEN CONNECTED TO A 6V BATTERY?
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n air-filled parallel-plate capacitor has plates of area 2.30 cm2 separated by 1.50 mm. (a) Find the value of its capacitance. The capacitor is connected to a 12.0-V battery. (b) What is the charge on the capacitor?
To calculate the capacitance of the given parallel plate capacitor, you can use the formula:
C = (ε₀ * A) / d
where:
C is the capacitance,
ε₀ is the permittivity of free space (ε₀ ≈ 8.854 x 10⁻¹² F/m),
A is the area of each aluminum plate,
d is the distance between the plates.
Given values:
Area, A = 2 x 10⁻⁴ m²
Distance, d = 0.03 mm = 0.03 x 10⁻³ m
Permittivity of free space, ε₀ = 8.854 x 10⁻¹² F/m
1. Calculate the capacitance:
C = (ε₀ * A) / d
C = (8.854 x 10⁻¹² F/m * 2 x 10⁻⁴ m²) / (0.03 x 10⁻³ m)
C = (8.854 x 10⁻¹² F/m x 2 x 10⁻⁴ m²) / (3 x 10⁻⁵ m)
C = 5.902666667 x 10⁻¹¹ F
Therefore, the capacitance of the given capacitor is approximately 5.9 x 10⁻¹¹ F.
2. To calculate the charge stored on the plates when connected to a 6V battery, you can use the formula:
Q = C * V
where:
Q is the charge,
C is the capacitance,
V is the voltage.
Given values:
Capacitance, C = 5.9 x 10⁻¹¹ F
Voltage, V = 6V
2. Calculate the charge:
Q = C * V
Q = 5.9 x 10⁻¹¹ F * 6V
Q = 3.54 x 10⁻¹⁰ C
Therefore, the charge stored on the plates of the capacitor when connected to a 6V battery is approximately 3.54 x 10⁻¹⁰ C.