Question
Asked by:
$1.00 MAgnitude
- From Physics: General-Physics
- Closed, but you can still post tutorials
- Due on Jul. 06, 2009
- Asked on Jul 03, 2009 at 10:31:49PM
Q:
In a uniform electric field, the potential changes from 14.0 V to 32.6 V in a displacement of 0.130 m (the direction is opposite to the field, as shown). What is the magnitude of the acceleration of a proton (charge 1.60×1019 C and mass 1.67×1027 kg) that is placed in this field?
a. Call A the beginning point of the displacement in the figure A, and the end of the displacement, B. What is the potential energy of an electron at point A and point B?
b. What is the magnitude of the electric field in the figure?
c. What is the force on the electron? In which direction is it? Is it the same at point A and at point B? Why?
d. The electron starts from rest at the beginning of the displacement. Argue why the electron will move in the direction of the displacement.
e. Use the Work – Kinetic Energy theorem (from Physics I) to find the speed of the electron at the end of the displacement. (You MUST use the Work – Kinetic Energy theorem.
