Electric Potential Change - General Physics - Solved Past Paper, Exams of Physics

Download Electric Potential Change - General Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! 3. ( pts) An electron moving to the right parallel to the x-axis has an initial speed of 1.4×105 m/s. After moving 3 cm through a uniform electric field, its final speed is 3.7 × 106 m/s. a) What is the difference in electric potential energy of the electron between the starting and ending points of this portion of its motion? (me = 9.11 × 10 −31 kg) ∆U = −∆K = ( 1 2 ) ( 9.11 × 10−31 ) [ ( 3.7 × 106 )2 − ( 1.4 × 105 )2 ] = −6.23 × 10−18 J b) By how much does the electric potential change between the starting and ending points? ∆V = ∆U q = −6.23 × 10−18 −1.6 × 10−19 = +38.9 V c) In what direction is the electric field? There are (at least) two ways to approach this problem: 1. V is increasing as the electron travels to the right, thus the electric field must point in the oppo- site direction, towards decreasing electric potential: to the left. 2. The electron, a negative charge, is experiencing a force to the right. Since it is a negative charge, the electric field (in order to produce the force to the right on the negative charge) must be to the left. d) What is the magnitude of the electric field? The problem specifies that it is a uniform electric field. Thus ∆V = − ∫ ~E · d~ℓ = −E d So in this case we end up with |E| = V/d = 38.9/0.03 = 1297 V/m