The answers to your questions are as written below
A) capacitance of a parallel plate capacitor = 442.5 * 10⁻¹³ F
B) The charge on each plate is = 2655 * 10⁻¹³ C
C) The electric field between the plates is ; 2 * 10³ N/C
D) The estimated area of the plates is ; 1.695 * 10⁹ m²
A) Calculate the capacitance of a parallel plate capacitor
Plate dimension = 30 cm * 5 cm
Air gap ( d ) = 3 mm = 3 * 10⁻³ m
First step : calculate the area of the plate
Area of plate = 30 * 10⁻²m * 5 * 10⁻² m
= 150 * 10⁻⁴ m²
Next step : determine the capacitance
Capacitance = ε₀A / d
= ( 8.85 * 10⁻¹² * 150 * 10⁻⁴ ) / 3 * 10⁻³
= 442.5 * 10⁻¹³ F
B) Calculate the charge on each plate if a 6-V battery is connected
charge on each plate can be calculated with the formula below
Q = CV
= 442.5 * 10⁻¹³ * 6
= 2655 * 10⁻¹³ C
C) Determine the electric field between the plates
Electric field between plates ( E ) = V / d
= 6 / 3 * 10⁻³
= 2 * 10³ N/C
D) Estimate the area of the plates
Required capacitance = 5 F
Air gap ( d ) = 3 * 10⁻³ m
applying the equation below
C = ε₀A / d -----
5 = ( 8.85 * 10⁻¹² * A ) / 3 * 10⁻³
Therefore A :
A = ( 5 * 3 * 10⁻³ ) / 8.85 * 10⁻¹²
= 15 * 10⁻³ / 8.85 * 10⁻¹²
= 1.695 * 10⁹ m²
Hence we can conclude that the answers to your questions are as written above.
Learn more about Capacitance of a parallel plate capacitor : https://brainly.com/question/17018704
