A 12.0- A 12.0- A 12.0-V battery remains connected to a parallel plate capacitor with a plate area of 0.204m^2 and a plate separation of 3.14mm .
What is the charge on the capacitor?
Q = C
How much energy is stored in the capacitor?
Uc = J
What is the electric field between its plates?
E = V/m
Calculate the capacitance (C = eo A/d)
then
Q = C V
and
energy = .5 Q V = .5 C V^2
then indeed
E = V/ spacing in meters between plates.
To find the charge on the capacitor, we need to use the formula Q = C, where Q is the charge and C is the capacitance.
To find the capacitance, we can use the formula C = ε₀ * (A / d), where ε₀ is the permittivity of free space, A is the plate area, and d is the plate separation.
Given:
Plate area (A) = 0.204 m^2
Plate separation (d) = 3.14 mm = 0.00314 m
First, let's calculate the capacitance:
C = ε₀ * (A / d)
From the given information, we need to know the value of ε₀, which is the permittivity of free space. ε₀ has a constant value of approximately 8.85 x 10^-12 F/m.
Using this value, we can calculate the capacitance:
C = (8.85 x 10^-12 F/m) * (0.204 m^2 / 0.00314 m)
Next, we can calculate the charge on the capacitor:
Q = C
Now, to find the energy stored in the capacitor, we can use the formula:
Uc = (1/2) * C * V^2, where Uc is the energy, C is the capacitance, and V is the voltage.
Given:
Voltage (V) = 12.0 V
Using the calculated capacitance, we can find the energy:
Uc = (1/2) * C * V^2
Finally, to find the electric field between the plates, we can use the formula:
E = V / d, where E is the electric field, V is the voltage, and d is the plate separation.
Using the given values, we can calculate the electric field:
E = V / d
By substituting the given values into the respective formulas, you can find the charge on the capacitor, energy stored in the capacitor, and the electric field between the plates.