A uniform, 2.00 × 106 -V/m electric field is used to accelerate an electron. (a) Calculate the energy the electron acquires once it has moved 40.0 cm across this field. (b) Find the distance it must move across the field to increase its energy by 50.0 GeV.

Consider a charge of size +3.8 x 10-4 C and mass 8 kg is traveling to the left towards a +1.7 x 10-4 C charge with speed 119 m/s.  The +1.7 x 10-4 C charge is so massive that it does not recoil in response to the repulsion of the approaching charge.  Calculate how close the two charges get, in m.  Use k = 9 x 109 N m2 / kg2. (Please answer to the fourth decimal place - i.e 14.3225)

Suppose an electron (q = -e = -1.6 x 10¬9 C,m=9.1 x 10¬3' kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for %3D the final speed of the electron. Express numerical answer in two significant figures. The potential energy U is related to the electron charge (-e) and potential Vab is related by the equation: U = Assuming all potential energy U is converted to kinetic energy K, K+U = 0 K = -U 1 Since K mv and using the formula for potential energy above, we arrive at an equation for speed: 2 v = ( 1/2 Plugging in values, the value of the electron's speed is: x 107 m/s V=

Calculate the speed of a proton and an electronic after each particle accelerates from rest through a potential difference of 355V.

A proton (m=1.67 x 10^-27 kg) travels a distance of 4.3 cm parallel to a uniform electric field 3.5 x 10^5 V/m between the plates shown in the figure. If the initial velocity is 3.2 x 10^5 m/s, find the magnitude of its final velocity in m/s. (Ignore gravity)

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Answers: a) 3.3 C c/m² 6. Two electrons are fixed 2.00 cm apart. Another electron is shot from infinity and comes to rest midway between the two. What was its initial speed? Answer: 318 m/s

An electron initially at rest, is allowed to accelerate through a potential difference of 1 V, gaining kinetic energy KEe, whereas a proton, also initially at rest, is let accelerate through a potential difference of 1 V, gaining kinetic energy KEp. As ∣qe∣ = ∣qp∣ but mp >> me, therefore, Group of answer choices KEe >> KEp KEe << KEp KEe = KEp All we can say, is KEe ≠  KEp

An electron is to be accelerated in a uniform electric field having a strength of 1.90  106 V/m. (a) What energy in keV is given to the electron if it is accelerated through 0.810 m?_________ keV(b) Over what distance would it have to be accelerated to increase its energy by 50.0 GeV (as in the Stanford Linear Accelerator, which is actually smaller than this)? ________km

10. An electron is accelerated from rest through a potential difference of 2.5 x 10° V. (a) What is the relativistic kinetic energy of the electron? (b) What is the speed of the electron? (c) What is the total relativistic energy of the electron? Given that the magnitude of the charge of the electron is 1.602 × 10-19 C and the mass of the electron is 9.11 × 10-3! kg.