Two 2.4 cm diameter disks spaced 1.5 mm apart form a parallel-plate capacitor. The electric field between the disks is 4.3×10^5 V/m.

What is the voltage across the capacitor?
How much charge is on each disk?
An electron is launched from the negative plate. It strikes the positive plate at a speed of 2.3×10^7 m/s. What was the electron's speed as it left the negative plate?

E x d is the voltage

d is the plate separation

Use Gauss' Law or Q = CV for the charge

The change in kinetic energy equals e V

To find the voltage across the capacitor, we can use the formula:

V = Ed

Where:
V = voltage
E = electric field
d = distance between the plates

Given:
E = 4.3×10^5 V/m
d = 1.5 mm = 0.0015 m

Substituting the given values into the formula, we can calculate the voltage across the capacitor:

V = (4.3×10^5 V/m) * (0.0015 m)
V = 645 V

Therefore, the voltage across the capacitor is 645 V.

To find the charge on each disk, we can use the formula:

Q = CV

Where:
Q = charge
C = capacitance
V = voltage

The capacitance of a parallel-plate capacitor is given by:

C = ε₀A/d

Where:
C = capacitance
ε₀ = permittivity of free space (8.85 x 10^-12 F/m)
A = area of the plates
d = distance between the plates

Given:
diameter of the disk = 2.4 cm = 0.024 m
radius of the disk = 0.012 m

Substituting these values into the formula for capacitance:

C = (8.85 x 10^-12 F/m) * (π * (0.012 m)^2) / 0.0015 m
C = 2.25 x 10^-10 F

Now, we can calculate the charge on each disk:

Q = (2.25 x 10^-10 F) * (645 V)
Q = 1.45 x 10^-7 C

Therefore, the charge on each disk is 1.45 x 10^-7 C.

To find the initial speed of the electron when it left the negative plate, we can use the principle of conservation of energy:

1/2 mv₁² = qV

Where:
m = mass of the electron = 9.11 x 10^-31 kg
v₁ = initial speed of the electron
q = charge of the electron = 1.602 x 10^-19 C
V = voltage across the capacitor = 645 V

Substituting the given values into the equation:

1/2 (9.11 x 10^-31 kg) v₁² = (1.602 x 10^-19 C) * (645 V)

Solving for v₁:

v₁² = (2 * (1.602 x 10^-19 C) * (645 V)) / (9.11 x 10^-31 kg)
v₁ = √((2 * (1.602 x 10^-19 C) * (645 V)) / (9.11 x 10^-31 kg))

Calculating the value of v₁:

v₁ = 3.25 x 10^6 m/s

Therefore, the initial speed of the electron when it left the negative plate was 3.25 x 10^6 m/s.

To find the voltage across the capacitor, we can use the formula for electric field between parallel plates:

Electric field (E) = Voltage (V) / Distance between plates (d)

Rearranging the formula, we can solve for voltage:

Voltage (V) = Electric field (E) × Distance between plates (d)

Plugging in the values for electric field (E = 4.3×10^5 V/m) and distance between plates (d = 1.5 mm = 0.0015 m), we can calculate the voltage:

V = (4.3×10^5 V/m) × (0.0015 m) = 645 V

Therefore, the voltage across the capacitor is 645 V.

To find the charge on each disk, we can use the formula for the capacitance of a parallel-plate capacitor:

Capacitance (C) = (Permittivity of free space) × (Area of each plate) / Distance between plates

The area of each plate can be calculated using the formula for the area of a circle:

Area (A) = π × (Radius)^2

Since the diameter of each disk is given as 2.4 cm, the radius would be half of that, which is 1.2 cm = 0.012 m.

Plugging in the values for the permittivity of free space (ε₀ = 8.85×10^-12 F/m), area of each plate (A = π × (0.012 m)^2), and distance between plates (d = 0.0015 m), we can calculate the capacitance:

C = (8.85×10^-12 F/m) × (π × (0.012 m)^2) / (0.0015 m) = 0.000357 F (or 3.57×10^-4 F)

The charge on each plate can be calculated using the formula:

Charge (Q) = Capacitance (C) × Voltage (V)

Plugging in the values for capacitance (C = 3.57×10^-4 F) and voltage (V = 645 V), we can calculate the charge:

Q = (3.57×10^-4 F) × (645 V) = 0.23085 C (or 2.3085×10^-1 C)

Therefore, the charge on each disk is approximately 0.231 C (or 2.31×10^-1 C).

To find the electron's speed as it left the negative plate, we can equate the kinetic energy gained by the electron to the change in electric potential energy.

The kinetic energy gained is given by:

Kinetic Energy (KE) = 1/2 × mass × (final velocity)^2

The change in electric potential energy is given by:

Change in Electric Potential Energy (ΔPE) = Charge (Q) × Voltage (V)

Since the electron is negatively charged, it will gain potential energy as it moves from the negative plate to the positive plate, so ΔPE will be positive.

Equating the two expressions, we get:

1/2 × mass × (final velocity)^2 = Charge (Q) × Voltage (V)

We are given the final velocity as 2.3×10^7 m/s and the voltage as 645 V. The charge on the electron is 1.6×10^-19 C, and the mass of the electron is 9.11×10^-31 kg.

Plugging in these values, we can solve for the initial velocity of the electron:

1/2 × (9.11×10^-31 kg) × (2.3×10^7 m/s)^2 = (1.6×10^-19 C) × (645 V)

Simplifying the equation, we find:

Initial velocity^2 = (2 × (1.6×10^-19 C × 645 V)) / (9.11×10^-31 kg)

Taking the square root of both sides, we can find the initial velocity of the electron:

Initial velocity = √((2 × (1.6×10^-19 C × 645 V)) / (9.11×10^-31 kg))

Calculating the square root, we get:

Initial velocity ≈ 3.32×10^6 m/s

Therefore, the electron's speed as it left the negative plate was approximately 3.32×10^6 m/s.