A certain parallel capacitor consist of two identical aluminium plates, each of area 2times 10^-4m^2. the plates are separated by a distance of 0,03mm, with air occupying the space between the plates. CALCULATE THE CAPACITANCE OF THE CAPACITOR. AND ALSO CALCULATE THE CHARGE STORED ON THE PLATES OF THE CAPACITOR WHEN CONNECTED TO A 6V BATTERY.
think Hard!!
To calculate the capacitance of the parallel capacitor, you can use the formula:
C = (ε₀ * A) / d
Where:
- C is the capacitance in farads (F)
- ε₀ is the permittivity of free space, approximately 8.85 × 10⁻¹² F/m
- A is the area of one plate in square meters (m²)
- d is the separation distance between the plates in meters (m)
Given values:
- A = 2 × 10⁻⁴ m² (area of each plate)
- d = 0.03 mm = 0.03 × 10⁻³ m (plate separation)
Let's calculate the capacitance:
C = (8.85 × 10⁻¹² F/m * 2 × 10⁻⁴ m²) / (0.03 × 10⁻³ m)
C = (8.85 × 2) / 0.03 × 10⁻⁶ μF (convert to microfarads)
C = 0.591 μF (rounded to three decimal places)
So, the capacitance of the capacitor is approximately 0.591 μF.
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 in coulombs (C)
- C is the capacitance in farads (F)
- V is the voltage across the capacitor in volts (V)
Given values:
- C = 0.591 μF (capacitance)
- V = 6V (voltage of the battery)
Let's calculate the charge:
Q = 0.591 μF * 6V
Q = 3.546 μC (rounded to three decimal places)
So, the charge stored on the plates of the capacitor when connected to a 6V battery is approximately 3.546 μC.