How Do You Calculate Electron Speed and Particle Charge in Physics Problems?

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To calculate the speed of an electron accelerated through a potential difference, the relationship between work and kinetic energy is essential. The formula W = qΔV relates the work done to the charge and potential difference, while the change in kinetic energy is expressed as W = ΔK = 1/2 m(v_f^2 - v_i^2). For the first problem, using the mass of the electron and the potential difference, the final speed can be derived. In the second question, the charge of a positively charged particle can be determined using its mass and achieved speed after acceleration through a potential difference. The elementary charge, e = 1.602 x 10^-19 C, is crucial for these calculations.
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Please help With 2 "Electric" Questions Please

I don't want to bother you guys, but my little brother has an independent study unit due tomorrow and he can't seem to figure out these two questions. I'm not so great with physics since I'm taking biology related courses. Anyways the questions are as follows, i think they are actually very simple, but i just don't have the background to figure them out right now. So if anyone can post solutions to these, it'll help myself learn/understand it so i can relay it to my brother. thanks.

1. An electron has a mass of 9.11 X 10^-31 kg. Originally, it's at rest. What speed will the electron have after it is accelerated through a potential difference of 5.67 X 10^3 V?

2. A positively charged particle had a mass of 6.70 X 10^-27 kg. When it was accelerated through a potential difference of 4.50 X 10^3 V, it achieved a speed of 6.56 X 10^5 m/s. What charge did the particle have?

Thanks again =).
 
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They are energy problems. Potential difference and work are related through

W = q \Delta V

And work is related to change of kinetic energy through

W = \Delta K = \frac{1}{2}m(v_f^2-v_1^2)
 
hmm, interesting, based on the info that you have given, I've deduced that the answer to #1 can be obtained through:

q/(4.555X10^-31) = vf^2

where q is a charge in coulombs, and vf is the final velocity.

how do i figure what q is?
 
\frac{2q\Delta V}{m} = v_f^2

I get q(1.2447 x 10^-28) = vf^2.

The charge q is that of the electron. It is the elementary charge e = 1.602 x 10^-19 C.
 
your info helped a lot thanks.
 
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