Question

A parallel-plate capacitor is formed from two 6.0-cm-diameter electrodes spaced 2.0 mm apart. The electric field strength inside the capacitor is 1.0*10^6 N/C. What is the charge (in nC) on each electrode?

Please show -all- steps for the derivation. The correct answer should be 25 nC and -25 nC

Answer 1

Answer 2

To find the charge on each electrode of the parallel-plate capacitor, we can use the formula for the electric field strength (E) between the plates of a capacitor:

E = σ / ε₀

where: E = Electric field strength (given as 1.0 * 10^6 N/C) σ = Surface charge density (charge per unit area) on the capacitor plates ε₀ = Permittivity of free space (a constant value: 8.85 * 10^-12 C^2/(N*m^2))

The surface charge density (σ) is the charge (Q) on one electrode divided by the area (A) of one electrode:

σ = Q / A

The area of one electrode (A) is the area of a circle with a diameter of 6.0 cm:

A = π * (diameter/2)^2

Now, let's solve for the charge on each electrode (Q):

Step 1: Calculate the area of one electrode (A):A = π * (6.0 cm / 2)^2 A = π * (3.0 cm)^2 A ≈ 28.27 cm²

Step 2: Calculate the surface charge density (σ):σ = Q / A σ = Q / 28.27 cm²

Step 3: Use the electric field strength formula to find Q:E = σ / ε₀ Q = E * A * ε₀

Step 4: Convert the charge from coulombs to nanocoulombs:1 C = 1 * 10^9 nC

Now, let's plug in the values:

Q = (1.0 * 10^6 N/C) * (28.27 cm²) * (8.85 * 10^-12 C^2/(N*m^2)) * (1 * 10^9 nC / 1 C)

Q ≈ 25 nC

The charge on each electrode is approximately 25 nC. Since the charge on the other electrode is equal in magnitude but opposite in sign, the total charge is -25 nC. Thus, the correct answer is 25 nC and -25 nC, as you mentioned.