An electric field of 850 kV/m is desired between two parallel plates, each area 7cm x 5 cm and separated by 2.5 mm of air.
a) what charge must be on each plate ?
b) how much energy is stored ?
To find the answers to these questions, we need to use the formulas related to electric fields, charges, and the energy stored in a capacitor.
a) To determine the charge on each plate, we can use the formula:
Q = Epsilon * A * V
Where:
Q = Charge on each plate
Epsilon = Permittivity of air (8.854 x 10^-12 F/m)
A = Area of each plate (in square meters)
V = Voltage (electric field) between the plates
Given values:
E = 850 kV/m (convert to V/m: 850 x 10^3 V/m)
A = 7 cm x 5 cm (convert to m^2: 7 cm x 5 cm = 0.07 m x 0.05 m = 0.0035 m^2)
Now, we can substitute the values into the formula:
Q = (8.854 x 10^-12 F/m) * (0.0035 m^2) * (850 x 10^3 V/m)
Calculating this equation will give you the charge on each plate in coulombs.
b) To determine the energy stored in the capacitor, we can use the formula:
E = (1/2) * C * V^2
Where:
E = Energy stored in the capacitor
C = Capacitance of the capacitor
V = Voltage (electric field) between the plates
To find the capacitance, we can use the formula:
C = Epsilon * A / d
Where:
C = Capacitance of the capacitor
Epsilon = Permittivity of air (8.854 x 10^-12 F/m)
A = Area of each plate (in square meters)
d = Separation between the plates
Given values:
A = 0.0035 m^2 (from part a))
d = 2.5 mm (convert to meters: 2.5 mm = 0.0025 m)
Now, we can calculate the capacitance:
C = (8.854 x 10^-12 F/m) * (0.0035 m^2) / (0.0025 m)
Once we have the capacitance, we can substitute it back in the energy formula to find the energy stored in the capacitor.
Remember to convert the units in all calculations to be consistent (such as meters for distance and F/m for permittivity) to get accurate results.