Question
Inside most old fashioned ("tube") TV sets you will find two parallel charged plates designed to accelerate electrons. Let's assume an electron (mass m = 9.11×10-31 kg) is accelerated in the uniform field E (E = 2.10×103 N/C) between the plates shown in the figure. (The thickness of the plates themselves is negligible, despite how fat they might look in this picture) The electron starts from rest near the negative plate and passes through a tiny hole in the positive plate, as seen in the figure below. It exits with a kinetic energy of 4.87×10-18 J (which, as we will see later in this course, is enough to generate light when it finally hits the phosphors on the front screen!) Inside most old fashioned ("tube") TV sets you will find two parallel charged plates designed to accelerate electrons. Let's assume an electron (mass m = 9.11×10-31 kg) is accelerated in the uniform field E (E = 2.10×103 N/C) between the plates shown in the figure. (The thickness of the plates themselves is negligible, despite how fat they might look in this picture) The electron starts from rest near the negative plate and passes through a tiny hole in the positive plate, as seen in the figure below. It exits with a kinetic energy of 4.87×10-18 J (which, as we will see later in this course, is enough to generate light when it finally hits the phosphors on the front screen!) In designing this system, how far apart do you need to space the two plates?
Explanation / Answer
Hi, The force acting on the electron due to the electric field = F = q * E = 1.6 * 10^(-19) * 2.10 * 10^3 = 3.36 * 10^(-16) N. Hence the acceleration on the electron = a = F / m = 3.36 * 10^(-16) / ( 9.11 * 10^(-31) ) = 0.3688 * 10^(15) m/s^2. Given the K.E at the point of exit from the capacitor = 4.87 * 10^(-18) J. => (1/2) * m * v^2 = 4.87 * 10^(-18) => v = 3.269 * 10^6 m/s. We know the electron started from rest at the first plate, i.e., u = 0. Hence from from v = u + a t we have 3.269 * 10^6 = 0.3688 * 10^(15) * t => t = 8.86 * 10^(-9) s. i.e., the electron should travel for a time 8.86 * 10^(-9) sec in the electric field to attain the said K.E. In this time t, it will have convered a distance s, which will be the distance between the plates. Hence from s = ut + (1/2) a t^2 we have s = 0.5 * 0.3688 * 10^(15) * [8.86 * 10^(-9)]^2 = 14.475 * 10^(-3) m Hope this helps you. Hope this helps you.