Two 1.50cm -diameter disks spaced 1.80mm apart form a parallel-plate capacitor. The electric field between the disks is 5.10×10^5 V/m
I got the voltage as =918
but
B. How much charge is on each disk?
_________C?
and
C. An electron is launched from the negative plate. It strikes the positive plate at a speed of 2.50×107m/s . What was the electron's speed as it left the negative plate?
______ m/s?
thank you so much guys!!
To find the charge on each disk, we can use the formula for the electric field between parallel plates:
E = V/d
where E is the electric field, V is the voltage, and d is the separation between the plates.
Given that the electric field is 5.10×10^5 V/m and the separation between the plates is 1.80 mm (or 1.80×10^-3 m), we can rearrange the formula to solve for V:
V = E * d
V = (5.10×10^5 V/m) * (1.80×10^-3 m)
V = 918 V
Therefore, your calculation of the voltage as 918 V is correct.
Now, to determine the charge on each disk, we use the formula for capacitance:
C = (ε₀ * A) / d
where C is the capacitance, ε₀ is the permittivity of free space (approximately 8.85×10^-12 F/m), A is the area of one disk, and d is the separation between the disks.
Given that the diameter of each disk is 1.50 cm (or 1.50×10^-2 m), we can calculate the area of one disk:
A = π * r^2
A = π * (d/2)^2
A = π * (1.50×10^-2 m/2)^2
A ≈ 0.001767 m^2
Substituting the values into the capacitance formula:
C = (8.85×10^-12 F/m) * (0.001767 m^2) / (1.80×10^-3 m)
C ≈ 8.69×10^-9 F
Since the charge on each disk is proportional to the voltage and capacitance, we can calculate it using the formula:
Q = C * V
Q = (8.69×10^-9 F) * (918 V)
Q ≈ 7.98×10^-6 C
Therefore, the charge on each disk is approximately 7.98×10^-6 C.
Moving on to the third question, to find the electron's speed as it left the negative plate, we need to use conservation of energy. The total mechanical energy of the electron is conserved, so we can equate the kinetic energy when it is launched from the negative plate with the kinetic energy when it strikes the positive plate.
The kinetic energy of an object is given by the formula:
KE = (1/2) * m * v^2
where KE is the kinetic energy, m is the mass of the object, and v is its velocity.
Assuming the electron's mass is 9.11×10^-31 kg, we can calculate its kinetic energy when it strikes the positive plate. Let's call this KE₂.
KE₂ = (1/2) * m * v₂^2
KE₂ = (1/2) * (9.11×10^-31 kg) * (2.50×10^7 m/s)^2
KE₂ ≈ 2.28×10^-15 J
Now, let's calculate the kinetic energy when the electron leaves the negative plate. Let's call this KE₁.
KE₁ = (1/2) * m * v₁^2
Since the electron leaves the negative plate under the influence of the electric field, its potential energy is converted into kinetic energy. The potential energy is given by the formula:
PE = q * V
where PE is the potential energy, q is the charge, and V is the voltage.
For the electron, the potential energy can be represented as:
PE = q * V = KE₁
Rearranging the formula to solve for the electron's velocity:
v₁ = sqrt(2 * q * V / m)
Substituting the known values:
v₁ = sqrt(2 * (7.98×10^-6 C) * (918 V) / (9.11×10^-31 kg))
v₁ ≈ 7.63×10^6 m/s
Therefore, the electron's speed as it left the negative plate is approximately 7.63×10^6 m/s.