Electrons emitted from the cathode of a cathode ray tube are accelerated by a potentialdifference of 2.00 kV between the cathode and the anode. The electrons then enter the uniformelectric field between the middle of two parallel conducting plates, each of length 2.00 cm. ThePotential difference between the plates is 500 V and the plates are separated by a distance of1.50 cm.Calculate(a) the velocity 𝑢 of the electrons when they emerge from the anode.(b) the time taken for the electrons to transverse the uniform electric field.(c) the velocity of the electrons when they emerge from the uniform electric field.
Question
Electrons emitted from the cathode of a cathode ray tube are accelerated by a potentialdifference of 2.00 kV between the cathode and the anode. The electrons then enter the uniformelectric field between the middle of two parallel conducting plates, each of length 2.00 cm. ThePotential difference between the plates is 500 V and the plates are separated by a distance of1.50 cm.Calculate(a) the velocity 𝑢 of the electrons when they emerge from the anode.(b) the time taken for the electrons to transverse the uniform electric field.(c) the velocity of the electrons when they emerge from the uniform electric field.
Solution
Sure, let's solve this step by step:
(a) The velocity 𝑢 of the electrons when they emerge from the anode.
We can use the equation for kinetic energy (KE = 1/2 mv^2) and equate it to the electric potential energy (qV), where m is the mass of the electron, v is the velocity, q is the charge of the electron, and V is the potential difference.
So, 1/2 mv^2 = qV
Rearranging for v, we get v = sqrt(2qV/m)
The charge of an electron (q) is 1.6 x 10^-19 C, the mass of an electron (m) is 9.11 x 10^-31 kg, and the potential difference (V) is 2000 V (or 2.00 kV).
Substituting these values in, we get v = sqrt(2*1.6 x 10^-19 C * 2000 V / 9.11 x 10^-31 kg) = 2.64 x 10^7 m/s
(b) The time taken for the electrons to transverse the uniform electric field.
The time taken can be calculated using the equation t = d/v, where d is the distance and v is the velocity.
The distance (d) is the length of the plates, which is 2.00 cm or 0.02 m. The velocity (v) is what we calculated in part (a).
So, t = 0.02 m / 2.64 x 10^7 m/s = 7.58 x 10^-10 s
(c) The velocity of the electrons when they emerge from the uniform electric field.
The velocity of the electrons will not change as they pass through the uniform electric field because the electric field only changes the direction of the electrons, not their speed. So, the velocity of the electrons when they emerge from the uniform electric field is the same as when they entered, which is 2.64 x 10^7 m/s.
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