I have tried this problem a couple of different ways using the resources given t
ID: 2002638 • Letter: I
Question
I have tried this problem a couple of different ways using the resources given to me and I just can't seem to figure it out. The potential difference given in the problem is 25000V so ignore on the information picture (3rd picture) the 25V. 2:13 PM edugen wileyplus.com iPad Step 1: Solution Step 1 Correct. The force that accelerates the electron is a conservative electric force, so that the total mechanical energy of the electron is conserved. Its kinetic energy increases as it accelerates from rest, so its electric potential energy must decrease. Note: Be aware that the numeric values in this stepped tutorial are different from the numeric values that appear in the question you Concept Questions. In a television picture tube, electrons strike the screen after being accelerated from rest through a potential difference. Ignoring the effects of special relativity, we are to determine the speed of the electrons just before they strike the screen. Consider a single electron being accelerated from rest. Which one of the following quantities must decrease during this process? The electron's kinetic energy. The electron's total mechanical energy. The electron's electric potential energy Step 2: Solution Step 2 Correct. Because the charge on the electron is negative, the fact that it freely accelerates from A to B indicates that the potential at A is lower than the potential at B. See Section 19.2. Note: Be aware that the numeric values in this stepped tutorial are different from the numeric values that appear in the question you are attempting to answer Concept Questions. Let the electric potential at the electron's initial position be denoted VA, and the electric potential at the position just before the electron strikes the screen be denoted Vs. Given that the electron has a negative charge, which one of the following is true?Explanation / Answer
c part : Potential Energy is given as U = kqQ/r = q(kQ/r) = q(potential difference)= q(VA-VB) = -e (25000V)
U=-25000eV= 25000/1.6e-19 J = 1.56e23 J