Electrostatics, finding velocity of proton

[physics604]

1. A moving proton has 6.4 x 10-16 J of kinetic energy. The proton is accelerated by a potential difference of 5 000 V between parallel plates. The proton emerges from the parallel plates with what speed?

a) 1.3 x 10⁶ m/s
b) 8.8 x 10⁵ m/s
c)1.8 x 10⁶ m/s
d) 9.8 x 10⁵ m/s

Homework Equations

$\Delta$E_k + $\Delta$E_p = 0

The Attempt at a Solution

E_ki + E_kf = -E_p
6.4×10^-16 + 1/2mv² = -Q$\Delta$v

v= √ [(-1.6x10^-19×-5000)-6.4x10^-16 ×2 / 1.67x10^-27] = 437740.5241...

My answer doesn't match with any of the responses. What did I do wrong?

 

[Simon Bridge]

Looks like you have misplaced a minus sign or three.
I suspect you have applied the formula without understanding it.

Derive the relation you need using:
Gain in kinetic energy = loss in potential energy

Hint:
what is the initial speed of the proton?
should the final speed be greater than or less than this?
what does this say about the final kinetic energy vs the initial kinetic energy?
how does this relate to the change in potential energy (careful)?

 

[physics604]

Okay, I get it!

$\Delta$E_k = E_kf - E_ki

I just used that part of the formula wrong.

 

[Simon Bridge]

Though had you gone to the physics first, you wouldn't have needed to know how to use any particular formula.
You can get a long way just looking for equation to stick the numbers you have into - but that way of thinking will always bite you eventually.

But "no worries" aye?