I think i did this electric potential problem correctly but

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SUMMARY

The problem involves a helium nucleus with a charge of 2e and mass of 6.63E-27 kg, traveling from an electric potential of 1.5E3 V to 4.0E3 V. The correct approach to find the speed at point B utilizes the equation delta V = - (delta KE)/q, leading to the conclusion that the nucleus slows down when moving against the electric field. The initial speed of 6.20E5 m/s results in a final speed of 3.78E5 m/s at point B, correcting the initial miscalculation of 7.91E5 m/s due to a sign error in the potential difference. The key takeaway is understanding the direction of electric fields and their impact on charged particles.

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Homework Statement



a helium nucleus (charge =2e, mass = 6.63E-27 kg) traveling at a speed of 6.20E5 m/s enters an electric field, traveling from point A, at a potential of 1.5E3 V, to point B, at 4.0E3 V. what is the speed at point B?

Homework Equations



charge = 2e = 2 * 1.6E-19 C
delta V is potential difference.
i used equation delta V = - (delta KE)/q and derived the relationship:
delta V = - mass/2q (vf^2 - vi^2)

The Attempt at a Solution


since the charge is 2e and e for electron is -1.6E-19 C, so the right hand side is positive. the left side is 4.0E3-1.5E3...finally i solved vf and got 7.91E5 m/s but the answer is 3.78E5 m/s

i don't get it...where is the mistake?
 
Last edited:
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I think you are making a sign error. Electric fields point from higher potential to lower potential. The helium nucleus is +2e (not a negative charge). This positive charge is going from A to B, or lower potential to higher potential. This means it is going against the E field. This means that the nucleus will slow down, not speed up. Notice that the answer you want is smaller than the initial velocity.

Note: I haven't done the math to check your equations. I'm just looking at what the answer, in general, should look like.
 

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