How many volts of potential difference is required to stop the electron with an initial velocity of 4.2 x10 ^ 5m / s? (qe = -1.6x10 ^ -19 c: m = 9.11x10-31 kg)

Two parallel metal plates separated by 20 cm are connected across a 12 V potential difference. An electron is released from rest at a location 10 cm from the negative plate. When the electron arrives at a distance 5 cm from the positive plate, how much kinetic energy has the electron gained? Hint: First you will need to calculate the value of the electric field between the plates, then the potential difference between the initial and final positions. Now use the Work–Energy Theorem. Standard units!

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.

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 electron (m = 9.1 x 10-31 kg, q = -1.6 x 10-1⁹ C) in the picture tube of a TV is accelerated from rest through a potential difference of AV = 5000 V. What is the final speed of the electron? CQ1: Which of the following is true? A) The electron accelerates from a region of lower potential to a region of higher potential (V, > V₁). B) The electron accelerates from a region of higher potential to a region of lower potential (Vƒ < Vj). C) There's not enough information to determine whether A or B is true.

The difference in electric potential between the accelerating plates in the electron gun of a TV tube is 25 kV, while the distance between them is 1.5 cm. The magnitude of the uniform electric field between the plates is 3.8 10 2 v/m. 5.6 10² v/m. O 1.7 106v/m. 11'108 v/m.

What is the magnitude of the electric field at the point (8.10i - 5.80j+9.40 k) m if the electric potential is given by V= 8.50xyz?, where Vis in volts and x, y, and z are in 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