A 3.00-cm-diameter parallel-plate capacitor with a spacing of 0.800mm is charged to 300V . What is the total energy stored in the electric field?

To calculate the total energy stored in the electric field of a parallel-plate capacitor, you can use the formula:

U = (1/2) * C * V^2

Where:
U is the energy stored in the electric field
C is the capacitance of the capacitor
V is the voltage across the capacitor

In this case, we are given the diameter of the parallel-plate capacitor and the spacing between the plates. We can use this information to calculate the capacitance of the capacitor, and then use the given voltage to find the total energy stored in the electric field.

First, let's find the capacitance of the capacitor:

The formula for the capacitance of a parallel-plate capacitor is:

C = (e0 * A) / d

Where:
C is the capacitance
e0 is the permittivity of free space (approximately 8.85 x 10^-12 F/m)
A is the area of the capacitor plates
d is the distance between the plates

Given that the diameter of the capacitor is 3.00 cm, the radius (r) can be calculated as:

r = diameter / 2 = 3.00 cm / 2 = 1.50 cm = 0.0150 m

The area (A) of the capacitor plates can be calculated as:

A = π * r^2

Substituting the values:

A = π * (0.0150 m)^2 ≈ 0.0007065 m^2

Now, we can calculate the capacitance (C):

C = (e0 * A) / d

The spacing between the plates is given as 0.800 mm, which can be converted to meters:

d = 0.800 mm = 0.800 x 10^-3 m = 0.000800 m

Substituting the values:

C = (8.85 x 10^-12 F/m * 0.0007065 m^2) / 0.000800 m ≈ 7.775 x 10^-12 F

Now, we can find the total energy stored in the electric field using the formula:

U = (1/2) * C * V^2

Substituting the capacitance (C) and the given voltage (V = 300V):

U = (1/2) * (7.775 x 10^-12 F) * (300 V)^2 ≈ 3.49875 x 10^-6 Joules

Therefore, the total energy stored in the electric field of the parallel-plate capacitor is approximately 3.49875 x 10^-6 Joules.