Ion Thruster Question from Year 12

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An ion thruster accelerates positive xenon ions by utilizing a potential difference between two grids, which increases their speed to 44 km/s. The mass of a xenon ion is 2.2x10^(-25) kg, and its charge is 1.6x10^(-19) C. To calculate the potential difference between Grid A and Grid B, one can apply the conservation of energy principle, equating the kinetic energy gained by the ion to the change in electrical potential energy. The discussion highlights the need for equations to solve the problem and acknowledges assistance from a user named astronuc. Understanding the relationship between kinetic energy and potential energy is crucial for solving the ion thruster's potential difference.
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Homework Statement


An ion thruster may be used to change the speed of a satellite, through the emmisions of charged particles. Positive xenons ions are accelerated by the potential difference(deltaV) in the region between grid A and grid B as they move towards the right of the diagram.

Xenon ions have a mass of 2.2x10^(-25) kg and a charge of 1.6x10^(-19) C

The potential diffrence in the ion thruster in the region between grid A and grid B increases the speed of an ion by 44 km^(-1)

Calculate the potential difference between Grid A and grid B. (assume that the ions enters the region between the grids with neglible speeds).

Homework Equations



we don't know!

The Attempt at a Solution



sure... with no equations? HELP!
 
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to the eight ppl who looked at it...yes we would like an answer
 
How about using conservation of energy. Assume a Xe ion starts at rest, and must be accelerated to a speed of 44 km/s. What is the kinetic energy of the ion at the exit? What is the change in electrical potential energy of the ion?
 
thanks astronuc, helped a heap.
 
Thread 'Correct statement about size of wire to produce larger extension'

Thread 'Correct statement about size of wire to produce larger extension'

The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...
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