A) An electron is to be accelerated from a velocity of 3.50×106 m/s to a velocity of 9.50×106 m/s . Through what potential difference must the electron pass to accomplish this? B)Through what potential difference must the electron pass if it is to be slowed from 9.50×106 m/s to a halt

(a) Calculate the speed of a proton that is accelerated from rest through a potential difference of 12o v. km/s (b) Calculate the speed of an electron that is accelerated through the same potential difference. Mm/s

A negatively charged particle of mass 8.0 x 10-13 kg is traveling rightward between two plates separated by a distance d = 50 cm, as shown below. The particle is launched with a speed of 1.0 x 104 m/s at the left plate, and strikes the right plate at a speed of 5.0 x 10³ m/s. The left plate is at electric potential V₁ = +1.0 V, while the right plate is at electric potential V₂ = -5.0 V. Assume the electric field between the plates is uniform, and ignore the effect of gravity. 2 (a) (b) V₁ d 7 V₂ Does the potential energy of the particle increase or decrease as it moves from the left plate to the right plate? Is the electric field between the plates directed rightward or leftward? Justify your answers briefly. Show clearly that the electric field between the plates has magnitude 12 V/m (or 12 N/C). Then find the charge of the particle (with correct magnitude and sign). (Note: You may obtain the charge using either energy considerations or a "force and kinematics" approach.

A proton has a speed of 2.0 x 105 m/s at a point where the electrical potential is 1000 V. It moves through a point where the electric potential is 700 V. What is its speed at this second point? (e = 1.6 x 10-19 C, mproton = 1.67 x 10.27 kg) 1.6 x 105 m/s 7.8 x 105 m/s 07 1.3 x 105 m/s 2.6 x 105 m/s 3.1 x 105 m/s 2.8 x 105 m/s