In the figure below, an electron is moving rightward between two parallel charged plates separated by distance d = 2.40 mm. The particle is speeding up from an initial speed of 3620 km/s at the left plate.
｜ ｜
｜ ○→ ｜
｜ ｜
-70V -50V
What is its speed just as it reaches the plate at the right? ?km/s
To find the speed of the electron just as it reaches the plate at the right, we need to consider the electric potential difference between the two plates and use the principle of conservation of energy.
The electric potential difference between the two plates is the difference in voltage, which is given as -70V at the left plate and -50V at the right plate.
The change in electric potential energy (ΔPE) of the electron as it moves from the left plate to the right plate is given by the formula:
ΔPE = qΔV
where q is the charge of the electron and ΔV is the change in voltage.
The kinetic energy (KE) of the electron just as it reaches the right plate is given by:
KE = (1/2)mv^2
where m is the mass of the electron and v is its final speed.
According to the conservation of energy principle, the change in electric potential energy is equal to the change in kinetic energy:
ΔPE = ΔKE
Setting these two equations equal to each other, we have:
qΔV = (1/2)mv^2
Now, let's substitute the given values:
q = charge of an electron = -1.602 x 10^-19 C (Coulombs)
ΔV = difference in voltage = -70V - (-50V) = -70V + 50V = -20V
m = mass of an electron = 9.109 x 10^-31 kg (kilograms)
Plugging in these values into the equation, we can solve for the final speed v:
(-1.602 x 10^-19 C) * (-20V) = (1/2)(9.109 x 10^-31 kg)(v^2)
v^2 = (2 * (-1.602 x 10^-19 C) * (-20V)) / (9.109 x 10^-31 kg)
v^2 = 0.35214 x 10^12 m^2/s^2
Taking the square root of both sides:
v = √(0.35214 x 10^12 m^2/s^2)
Now, let's convert the speed to km/s:
v = √(0.35214 x 10^12 m^2/s^2) * (1 km/1000 m) * (1 s/1000 ms)
v ≈ 594 km/s
Therefore, the final speed of the electron just as it reaches the plate at the right is approximately 594 km/s.