Calculating Electron Acceleration in an X-ray Machine

AI Thread Summary
To calculate the acceleration of an electron in an x-ray machine's electric field of 6 x 10^5 N/C, the relationship between electric field, force, and charge is utilized. The force on the electron is determined using F = E * q, resulting in a force of 9.6 x 10^-14 N. This force is then used in Newton's second law, F = m * a, to find the acceleration. The calculated acceleration is approximately 1.05 x 10^17 m/s², which appears to be correct based on the provided values. The discussion emphasizes understanding the relationship between electric fields, force, and particle mass in calculating acceleration.
wakejosh
Messages
101
Reaction score
0

Homework Statement


In x-ray machines, electrons are subjected to electric fields as great as 6 x 10^5 N/C. Find an electron's acceleration in this field. ( me = 9.1 x 10^-31kg, qproton= 1.6 x 10^-19C)



The Attempt at a Solution




can someone just point me in the right direction to solve this? a site with information etc. would help. I don't have the correct book for the class yet and I can't seem to find the info I need in this old book. Thanks
 
Last edited:
Physics news on Phys.org
An electric field is the force per unit charge. What do you know about force and acceleration, and what do you know about the charge of an electron?
 
Well here is a hint. The acceleration depends on the strength of the electric field and the charge of the particle. Also the mass of the particle has something to do with it. Are these enough hints for you?

Edit: sorry for stepping in Kurdt. you replied whilst I was typing this up
 
Last edited:
ok so I use:

E = F / q (E= electronic field, F= force, q= charge)

6x10^5 = F / 1.6x10^-19

gives me:

F= 9.6x10^-14

then we use F=M*A to find the acceleration:

9.6x10^-14 / 9.1x10^-31 = A = 1.05x10^17

is this correct?
 
looks fine
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top