How to find the speed of an ion using V, T, m, q?

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To find the speed of a Li+ ion accelerated through a potential difference of 400V and entering a magnetic field of 0.5T, the kinetic energy gained by the ion can be calculated using the equation EPE = qV, resulting in a value of 6.4E-17 J. This energy is equal to the kinetic energy (KE) of the ion, given by KE = 0.5mv². By rearranging this equation, the speed (v) can be determined. The discussion also clarifies that gravitational potential energy is not relevant in this scenario, as only electrical potential energy is considered. Understanding the conservation of energy principle is key to solving the problem effectively.
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Hello PF! How I've missed you! :)
Studying for a final that's in two days. Our prof skimped on the electricity/electron/unit thing. I don't even know what it's called! :(

Homework Statement



I need to find the speed of an ion, as well as the magnitude of the magnetic force acting on it.

A Li+ ion,

m= 1.16E-26 kg
q= 1.60E-19 C
V (or EPE? Same thing?)= 400V
v= ?
B= 0.5 T

Is accelerated through a potential difference of 400V and enters a uniform magnetic field of 0.5T, perpendicular to the ion's velocity. Find v of the ion, find magnetic force acting on the ion.

Homework Equations



I'm given the formulas (on a magical exam formula sheet)
q=Ne
EPE=qV
E=( kq) / r^2
F= ( kq1q2 ) /r^2


The Attempt at a Solution



F=qE but don't have E...
F= qvBsin0 but don't have v.

Suggestions? I have a horrendous understanding of this unit... Our entire class is actually lost because of the lack of explanation on it :(
 
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EPE is Electric Potential Energy. The units would be Joules. The potential difference through which the ion "falls" is 400V (Volts).

The unit "Volt" is equivalent to "Joules per coulomb". So if the ion with a charge of q= 1.60E-19 C falls through a 400V potential, what's the kinetic energy (Joules) it ends up with? Hint: you've listed an equation, EPE = qV, that is key (along with the relationship between work and energy).
 
gneill said:
EPE is Electric Potential Energy. The units would be Joules. The potential difference through which the ion "falls" is 400V (Volts).

The unit "Volt" is equivalent to "Joules per coulomb". So if the ion with a charge of q= 1.60E-19 C falls through a 400V potential, what's the kinetic energy (Joules) it ends up with? Hint: you've listed an equation, EPE = qV, that is key (along with the relationship between work and energy).

Still not seeing where you're going with this.. I'm sorry, I'm terrible.
EPE= qV
= 1.6E-19 x 400V
= 6.4E-17 J

KE= 0.5mv2...
 
Apollinaria said:
Still not seeing where you're going with this.. I'm sorry, I'm terrible.
EPE= qV
= 1.6E-19 x 400V
= 6.4E-17 J

KE= 0.5mv2...

Yes, you're doing fine. By conservation of energy, the KE gained by the "falling" charge equals the change in potential energy. KE = EPE.

So solve for the velocity.
 
gneill said:
Yes, you're doing fine. By conservation of energy, the KE gained by the "falling" charge equals the change in potential energy. KE = EPE.

So solve for the velocity.

Thats exactly what I suspected but, starting with this

KEi + PEi + EPEi = KEf + PEf + EPEf

where does everything else go to leave me with KE and EPE? Kinetic because its falling...

Edit: Got the correct answer but still curious as to where the other energies disappeared to.
 
Apollinaria said:
Thats exactly what I suspected but, starting with this

KEi + PEi + EPEi = KEf + PEf + EPEf

where does everything else go to leave me with KE and EPE? Kinetic because its falling...

Edit: Got the correct answer but still curious as to where the other energies disappeared to.

The problem statement makes no mention of gravity, so you're only dealing with Electrical Potential Energy here. So your PEi and PEf terms (presumably associated with gravitational potential) vanish. I used "falling" in quotes to denote that the charge is "falling" through an electrical potential difference, sorry if that was misleading :blushing:

Dropping the PE terms from your expression,

KEi + EPEi = KEf + EPEf

KEf - KEi = EPEi - EPEf

ΔKE = ΔEPE
 
gneill said:
The problem statement makes no mention of gravity, so you're only dealing with Electrical Potential Energy here. So your PEi and PEf terms (presumably associated with gravitational potential) vanish. I used "falling" in quotes to denote that the charge is "falling" through an electrical potential difference, sorry if that was misleading :blushing:

Dropping the PE terms from your expression,

KEi + EPEi = KEf + EPEf

KEf - KEi = EPEi - EPEf

ΔKE = ΔEPE

I apologize again for being so dense. Why is there a change in KE and EPE?
:confused:
I'm usually better at this. I don't know what's wrong with me today! :cry:
 
Apollinaria said:
I apologize again for being so dense. Why is there a change in KE and EPE?
:confused:
The ion is moving through a potential difference, so there's a change in EPE (which, as you may recall, is position dependent on the location in a field). The ion is changing speed as it does so, so there's a change in KE. The change in KE balances the change in EPE. That's conservation of energy at work. Works for any conservative field, such as gravitational or electric fields.
I'm usually better at this. I don't know what's wrong with me today! :cry:

Deep breaths, have some tea :smile:
 
gneill said:
The ion is moving through a potential difference, so there's a change in EPE (which, as you may recall, is position dependent on the location in a field). The ion is changing speed as it does so, so there's a change in KE. The change in KE balances the change in EPE. That's conservation of energy at work. Works for any conservative field, such as gravitational or electric fields.


Deep breaths, have some tea :smile:

Ooooooooooohhh. Now I get it! :approve: The more I look at these the more panicked I get. I have one day to finish my studies before the final :(
 
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