equation that relates voltage to kinetic energy/velocity?

[amar21]

does anyone know if such an equation exists?

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[pitt1166]

V=E*d maybe?

 

[HawkeyePostOp]

in V=E*d, E is field, not energy.

The potential energy on charge q from voltage V
PE = q*V
PE can then be converted to KE = (1/2)mv^2 if the problem needs it.

 

[minhaj]

EPE = qEd
KE = 1/2 mv^2

at the point of max velocity

qEd = 1/2qv^2

Vmax = (Ed)^.5

 

[domani]

wait, how does q=m at max velocity? i thought you needed q and m?
try this:

Work_total = qV = (1/2)m(v_max)^2
v_max = [(2qV)/m]^(0.5) or [(2qEd)/m]^(0.5)

 

[ballofnerves]

EPE = qEd
KE = 1/2 mv^2

at the point of max velocity

qEd = 1/2qv^2

Vmax = (Ed)^.5

This is incorrect.

wait, how does q=m at max velocity? i thought you needed q and m?
try this:

Work_total = qV = (1/2)m(v_max)^2
v_max = [(2qV)/m]^(0.5) or [(2qEd)/m]^(0.5)

This is correct.

 

[amar21]

thanks guys
i was just trying to put an equation into my mental database that shows energy increases as voltage increases.... it came up in CBT3 and I got it wrong, because I couldnt think of an equation that ever said they were related.

 

[ecoli]

alternatively, if its a question about a a charge moving in a magnetic feild, you can set
F = qvB = mv^2/r
which reduces to

qB = mv/r

 

[daemongray]

My issue is that I am confused as to how do you determine velocity when they tell you that the Voltage has been increased by a factor of four?

 

[QrtrLifeCrisis]

My issue is that I am confused as to how do you determine velocity when they tell you that the Voltage has been increased by a factor of four?

You would also need to know the charge.

Assuming all of the EPE is converted to KE:

q*V = EPE = KE = 1/2mv^2

You can solve for velocity.