In a uniform electric field, the potential changes from 14.0 V (starting point A
ID: 2025429 • Letter: I
Question
In a uniform electric field, the potential changes from 14.0 V (starting point A) to 32.6 V (ending point B) with a displacement of 0.130m (the direction is opposite to the field.)
What is the potential energy of an electron at point A and point B?
What is the magnitude of the electric field?
What is the force on the electron? In which direction is it? Is it the same at point A and at point B? Why?
The electron starts from rest at the beginning of the displacement. Argue why the electron should move in the direction of the displacement.
Use the Work – Kinetic Energy theorem (from Physics I) to find the speed of the electron at the end of the displacement. (W = ?KE)
NOTE: electric field (E) is moving to the left. the electron is moving opposite to that to the right.
Explanation / Answer
potential at point : A , VA = 14.0 V (starting point) potenial at point : B , VB = 32.6 V (ending point ) potentail difference b/w the points ,V = 32.6 V-14.0 V= 18.6 V charge on the electron , q = -1.6x10-19 C displacement of the charge , d = 0.130 m mass of the electron , m = 9.1 x10-31 kg (the direction is opposite to the field.)(a) when the charge moved from point A to pint B , an amount of work is done and which is stored as a potentail -energy in the electron . thus, work done = potential energy = q V = (-1.6x10-19 C)(18.6 V) = - 29.76 x10-19 J where , -ve sign indicates work is done against to the electric field (b) magnitude of the electric field , E = V /d = 18.6 / 0.130 = 143.07 V/m (c) force on the electron , F = Eq = (143.07)(-1.6x10-19) = -228.912 x10-19 N since force is negative , direction is point B to A (d) let , initial speed of the electron is , u = 0 m/s at the time of instant , let speed of the electron , v thus, change of the kinetic energy of electron is =1/2mv2 Acc. to work -energy therom change of the kinetic enrgy of the electron is equal to work done mathematically , Work done =W = 1/2mv2 29.76 x10-19 J= 1/2[(9.1x10-31 )][v]2 v = 2.557 x106 m/s thus, speed of the electron at the end of the displacement is , v = 2.557 x106 m/s v = 2.557 x106 m/s thus, speed of the electron at the end of the displacement is , v = 2.557 x106 m/s
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