In large CRT televisions, electrons are accelerated from rest by a potential difference of 23.88 kV and shot onto a phosphorescent screen to produce an image. What is the speed of the electrons when they reach the screen? ( qe = 1.602 x 10 19C ;mel = 9.11 x 10 31 kg) %3D Answer: x10' (express your answers in tenths place or one decimal digit only)

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=

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.

What is the magnitude of the work done if 3.0 C is transported between two points with a potential difference of 90 V?   a. 270 J b.30 J c.90 J   d.180 J

3a. What is the potential energy U of this lightning bolt? U = 310 Joules of the square? Find 30 (1.0x10) 3b. If all this energy where to be converted to kinetic energy, what would be the speed of a m = 1000 kg car. Vf = r16 = 34 = m/s 16. These charges are numbered as: raw the re Redraw the rectangle with the actual charges (evaluate 2q1, 492, etc.) 16a. What is the distance between the indicated pairs of charges. Find these in meters. r24 = r26 = 15 = 35 = 1 2 3 MI 4 5 6 ainT 16b. Starting with all the charges at ∞, what is the potential at the upper left corner, and how much work is done to put the charge at the upper left corner? V₁ = 05 P278 Volts W₁ = Joules V3 = V13 + V23 = | srpC = 10-12 C on m 16c. What is the potential at the location of 2, due to the previous charges? How much work is done to bring in the 2nd charge? Make sure you use the correct distance. V₂ = V12 = Volts W₂ = 16d. What is the potential at the location of 3, due to the previous charges? How much work is…

an evacuated tubes a potential differences of v=0.24 Kv to accelerate electrons which then hit a copper plate and produce x rays   Write an expression for the non-relativistic speed of these electrons v in terms of e, AV, and m, assuming the electrons start from rest

An electron moving parallel to the x axis has an initial speed of 3.7 x 106 m/s at the origin. Its speed is reduced to 1.4 x 105 m/s at x = 2 cm. Calculate the electric potential difference 1. between the origin and x -= 2 cm. 2. A proton is released from rest in a uniform electric field whose magnitude is 5000 V/m. Through what potential difference will it have passed after moving 0.25 meters? How fast will it be going after it has travelled 0.25 meters?

Find the potential at point P₂ in the diagram due to the two given charges. + +5 mC 4 cm- P₁ T 3 cm 2 cm 3 cm O a. 900MV O b. 0.8GV O C. -900MV O d. -4GV P3 4 cm- m 2 cm P2₂ -10 MC ←

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