If an electron is accelerated from rest through a potential differenceof 9.9 kV, what is its resulting speed? (e = 1.60 x 10-19 C. k= 1/4xe0%3D= 8.99 x 109 N-m2/C2, mel = 9.11 x 10-31 kg)%3DO 5.9 × 10' m/sO 3.9 x 107 m/sO 2.9 x 107 m/sO None of the given choices.O 4.9 x 107 m/s

Answered: Image /qna-images/answer/5639cac4-8683-48bc-8e12-be22a21c889e.jpg

If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed? (e = 1.60 × 10-19 C. k= 1/4xe0 = 8.99 x 109 N•m2/C2, mel = 9.11 x 10-31 kg) %3D O 5.9 x 10 m/s O 3.9 x 107 m/s O 2.9 x 107 m/s O None of the given choices. O 4.9 x 107 m/s

If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed? (e = 1.60 × 10-19 C. k= 1/4xe0 = 8.99 x 109 N•m2/C2, mel = 9.11 x 10-31 kg) %3D O 5.9 x 10 m/s O 3.9 x 107 m/s O 2.9 x 107 m/s O None of the given choices. O 4.9 x 107 m/s

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Related questions

Q: A charge of 9.3 nC is placed at the origin. Calculate the electric potential due to this charge at…

Q: What is the electric potential, in volts, due to the proton on an electron in an orbit with radius…

Q: An electron is accelerating under a potential difference of 105 volts. (m,=9,108x1031) a) What is…

A: The potential energy of the electron gets converted into the kinetic energy. Thus, the kinetic…

Q: A negative charge q= -9.0x 10-3 C is released from rest at point A, where the electric potential is…

Q: 1. A proton has a speed of 1.5 x 105 m/s at a point where the electrical potential is 1000 V. It…

Q: our answer in megajoules. QA Qc S (+)

Q: P d d d 92

Q: How many electrons must be removed from an uncharged spherical conductor of radius 0.300 m to…

A: The potential at the surface of a charged spherical conductor can be calculated using Gauss's law.…

Q: Charges Q1 = + 3.2 x 10-19 C and Q2 = - 4.8 x 10-19 C are located at x = 5.0 cm and x = 10.0 cm,…

A: Let potential be zero at x=x0

Q: total electric field E and the potential V at point P 5 (10,10,0)cm. Position vectors are given in…

Q: Two particles with Q1 = 48 µC and Q2 = 91 µC are initially separated by a distance of 2.2 m and then…

A: Given : Charges Q1 = 48 µC = 48 x 10-6 C and Q2 = 91 µC = 91 x 10-6 C Distance d1 = 2.2m ; Final…

Q: If a Cu2+ ion that is initially at rest accelerates through a potential difference of 12 V without…

Q: What is the potential energy of an electron at the point P of the capacitor shown in the figu 19 C)…

Q: Review Multiple-Concept Example 4 to see the concepts that are pertinent here. In a television…

A: potential difference = 22000 V

Q: An electron accelerates through a potential difference of 8.0 kV. How much kinetic energy does this…

A: Given: e=1.6×10-19 CV= 8 kV

Q: Two charges of q, = 1.6 µC and q, = -2.6 µC are d = 0.30 m apart at two vertices of an equilateral…

Q: q = -1.60 × 10-19 C and mass m = 9.11 × 10-³1 kg) has an initial speed Vo = 2.80 x 107 m/sec. The…

Q: A SYSTEM OF TWO ELECTRIC CHARGES b (in J) 92 91 What is the potential energy of this system of two…

Q: n the figure point P is at the center of the rectangle. With V - 0 at infinity, q1 = 4.70 fC, q2 =…

Q: The electric potential in a certain region is V = α x^2y^2 + β z^2(x^2 − γ) + δ y^3z , where α = 8…

A: Given: The Electric potential in a certain region V=αx2y2+βz2(x2-γ)+δy3zwhere:α=8 V/m4 , β=4.6 V/m4…

Q: The electrons in a CRT are accelerated through a potential difference of 30kV. Do the electrons move…

A: Given The potential difference is ∆V=30 kV=30×103 V. If the potential difference is positive then…

Q: What is the magnitude of the electric field at the point (2.80 i - 5.30 j +5.70 k) m if the electric…

A: Given: co-ordinates of the point= (2.80i^-5.30j^+5.70k^) m Electric potential V= 2.20xyz2…

Q: Across what potential difference must an electron, initially at rest, move to attain a speed of 1.0…

Q: A particle with a mass of 1.50x10^-5kg and a negative charge of -9.00x10^-6C is constrained to move…

A: Mass of particle = 1.5*10^-5kg Negative charge (q) = -9*10^-6C V(x) = 3x^4 volt

Q: An electron is accelerated from rest through a potential difference of 181 V. What is the electron's…

Q: (a)How fast is the proton moving (in m/s) when it is 3.50 meters away from thepoint charge? Use…

Q: If an electron is accelerated from rest through a potential difference of 1200 V, find its…

Q: 3. How much work must we do on an electron to move it from point A, which is at a potential of +50V,…

Q: Considering electron and proton as two charged particles separated by d = 4.5 × 10-11 m calculate…

Q: Review Multiple-Concept Example 4 to see the concepts that are pertinent here. In a television…

Q: Three charges q1 = -3.40 nC, q2 = 4.00 nC, q3 = 13.00 nC are separated by a distance L = 2.60 m.…

A: Given, three charges,q1 = -3.40 nCq2 = 4.00 nCq3 = 13.00 nCL = 2.60 m

Concept explainers

Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

Question

If an electron is accelerated from rest through a potential difference
of 9.9 kV, what is its resulting speed? (e = 1.60 x 10-19 C. k= 1/4xe0
%3D
= 8.99 x 109 N-m2/C2, mel = 9.11 x 10-31 kg)
%3D
O 5.9 × 10' m/s
O 3.9 x 107 m/s
O 2.9 x 107 m/s
O None of the given choices.
O 4.9 x 107 m/s

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions

(a) Find the electric potential difference AV required to stop an electron (called a "stopping potential") moving with an initial speed of 2.41 x 10 e m/s. kV (b) Would a proton traveling at the same speed require a greater or lesser magnitude of electric potential difference? O greater lesser O equal Explain. I
A proton is traveling in an electric field with a speed of 9.36 × 104 m/s. The proton slows down to a temporarily rest where the electric potential is measured to be 2.22 V. What is the electric potential at the initial position?
Review Multiple-Concept Example 4 to see the concepts that are pertinent here. In a television picture tube, electrons strike the screen after being accelerated from rest through a potential difference of 21000 V. The speeds of the electrons are quite large, and for accurate calculations of the speeds, the effects of special relativity must be taken into account. Ignoring such effects, find the electron speed just before the electron strikes the screen. VB =
so the distance is from -3 to 5 cm which is 8cm or .08m. I did parts a-e and need help for parts f-h. This is what I have so far: a: V1=1000V b: 3.2e^-16 J c. potential energy; no work needed to move particle as it leaves electric field d. 3.2e^-16 J e. m1<m2<m3<m4, v1>v2>v3>v4 f. distance =.08m -time takes particle 1 hit detector: .98e^-6 sec ---> .08m/.98e^-6 s= 8.16e^4 m/s -time takes particle 2 hit detector: 1.4e^-6 s --> .08m/1.4e^-6 s= 5.71e^4 m/s -time takes particle 3 hit detector: 1.7e^-6 s --> .08m/1.7^-6 s= 4.7053^4 m/s -time takes particle 4 hit detector: 1.96^-6 s --> .08m/1.96^-6 s= 4.082e^4 m/s I am not sure if i Made any mistakes so please let me know if i DO!
Considering electron and proton as two charged particles separated by d = 4.3 x 10-11 m calculate the electrostatic potential at a distance d from the proton. Take the mass of the proton 1.7 x 10-27 kg, the mass of the electron 9.1 x 10-31 kg, the electron charge -1.6 × 10-¹⁹ 1 C and = 9 × 10⁹ m/F. Απευ Answer: Choose...
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
In the figure shown below, q = 2.0 x 10-9 C and d = 1.2 m. The electric potential at point P located at the center of the square is: -2.0q -3.0q d- d •P d d- -2.09 +5.0q
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.
Two charges of q1 = 1.9 µC and q2 = −2.9 µC are d = 0.46 m apart at two vertices of an equilateral triangle as in the figure below. (a) What is the electric potential due to the 1.9-µC charge at the third vertex, point P? (in volts) (b) What is the electric potential due to the −2.9-µC charge at P? (in volts) (c) Find the total electric potential at P. (in volts) (d) What is the work required to move a 4.0-µC charge from infinity to P? (in joules)
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?
wot is the net electric potential at point P? 91=10.0nC 3,0m 4.0cm P 3.0em 92 - 12.0nc k=9.0x10"Nm? E- k lal e-1.6xloC tcom =10m V=k q
I+= 3.04 A 6031 15/60+ 10/60+ 1/60 1/3½60 7) The electric potential at points in the xy V= 2x²-3y² plane is X The electric field at the point (3m, 2m) 15 a)-6² +63 c) -12² + 123 Ex = -2v 2x Ey =-2v чу b) 122-125 4) 62-63 v=2x²-3y² z dv=4x-64 z Ex = − 2 (23) ²-3 (3) 18 = 3 2×(3) V=2x²-3y² 3.92 3í + 4.5g Ey = -2(2x=3y²) 18 = 4.5 2y 4 е) поне above of the