RaySix

Post

Capacitance and Energy in a Parallel Plate Capacitor

Created by @nathanedwards

at November 3rd 2023, 6:10:11 pm.

AP_Physics_1-Questions

#physics

#capacitors

#electric-fields

Question

A parallel plate capacitor has a plate separation of 0.02 meters and a plate area of 0.005 square meters. The capacitor is connected to a battery with a voltage of 12 volts.

a) Calculate the capacitance of the parallel plate capacitor.

b) When the battery is connected to the capacitor, a charge of 2 microcoulombs is stored on each plate. Calculate the electric field between the plates.

c) Calculate the stored energy in the capacitor.

Answer

a) The capacitance of a parallel plate capacitor can be calculated using the formula:

C = \frac{{\epsilon_0 \cdot A}}{{d}}

Where:

C is the capacitance of the capacitor
$\epsilon_0$ is the permittivity of free space (8.85 x 10^-12 F/m)
A is the area of the capacitor plates
d is the separation between the plates

Plugging in the given values:

C = \frac{{(8.85 \times 10^{-12} \, F/m) \cdot (0.005 \, m^2)}}{{0.02 \, m}}

Calculating:

C = 2.2125 \times 10^{-11} \, F

Therefore, the capacitance of the parallel plate capacitor is $2.2125 \times 10^{-11} \, F$ .

b) The electric field between the plates of a parallel plate capacitor can be calculated using the equation:

E = \frac{{V}}{{d}}

Where:

E is the electric field
V is the voltage
d is the separation between the plates

Plugging in the given values:

E = \frac{{12 \, V}}{{0.02 \, m}}

Calculating:

E = 600 \, V/m

Therefore, the electric field between the plates is $600 \, V/m$ .

c) The energy stored in a capacitor can be calculated using the equation:

U = \frac{{1}}{{2}} \cdot C \cdot V^2

Where:

U is the stored energy
C is the capacitance
V is the voltage

Plugging in the given values:

U = \frac{{1}}{{2}} \cdot (2.2125 \times 10^{-11} \, F) \cdot (12 \, V)^2

Calculating:

U = 7.9875 \times 10^{-10} \, J

Therefore, the stored energy in the capacitor is $7.9875 \times 10^{-10} \, J$ .