Mass of an accelerated electron?

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SUMMARY

An electron accelerated from rest through a potential difference of 31 kV gains energy calculated using the formula E = qV, where q is the charge of the electron (1.602 x 10^-19 C) and V is the potential difference. The mass of the electron at rest is m_0 = 9.109 x 10^-31 kg, and the relativistic mass m can be determined using m = m_0 + qV/c^2, resulting in a final mass of approximately 9.66 x 10^-31 kg. The discussion confirms that the energy gained during acceleration is correctly interpreted as the product of charge and voltage, reinforcing the relationship between energy and mass in relativistic physics.

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Homework Statement


An electron is accelerated from rest through a potential difference of 31kV

Homework Equations


I don't know how to engage or solve a problem like this, I've done this attempt but I'm very uncertain that I've done it in a correct way, it's hard to understand if I've choosen correct forumulas etc it feels much like guesswork for me, but the final answer seem plausible as it is a small increase as it should, shouldn't it?

Have I done it right and have a good solution? if not, how do you solve such a problem?

My feeling was that if there is a mass while rest and another for acceleration I can add them together like this. But I'm not sure I have the right masses even.

The Attempt at a Solution


the electrons mass, m_0 = 9.109*10^-31 kg
Speed of light, c = 2.998*10^8 m/s

E=m_1*c^2 and E=QU give me:
m_1 = QU / c^2

m_1 = (1.602*10^-19) * 31000 / (2.998*10^8)^2 = 5.525*10^-32 kg

I then add m_1 and m_0 together giving me my final answer. m, m= m_0 + m_1
m = (5.525*10^-32) + (9.109*10^-31) = 9.66*10^-31 kg

Very thankful for input :)
 
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This is OK! The electron gains energy qV , so total energy will be: E = m0c2 +qV = m c2 or m = m0 + qV/c2
 
when you say qV? is that QU? (columb times volt) or something else

so essentially I did what you describe in the end?
 
Tryhard2 said:
is that QU?
Yes

Tryhard2 said:
so essentially I did what you describe in the end?
Yes, I just referred the reasoning for doing this...
 

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