Potential difference between two plates

[moenste]

Homework Statement

An α-particle accelerated between a pair of parallel plates in a vacuum tube acquires a kinetic energy of 10³ eV. What is the potential difference between the plates?

Answer: 500 V

2. The attempt at a solution
Well, it's 1000 / 2 = 500 V because there are two plates. But I am not sure whether this is the right logic.

Also KE = e * V. If we use this, we have (1000 * 1.6 * 10^-19) J = 1.6 * 10^-19 * V and so V = 1000 V. And not 500 V as in the answer.

Is it correct to just divide 1000 by 2 and get 500 V and that is the answer in this case?

Thanks in advance!

 

[Hamal_Arietis]

It's wrong, two plates you can seem capacitor and between two plates have electric field
and it provides for paticles a KE. You must find the work is caused by electric field. It equals $\Delta KE$

 

[moenste]

It's wrong, two plates you can seem capacitor and between two plates have electric field
and it provides for paticles a KE. You must find the work is caused by electric field. It equals $\Delta KE$

Thank you. Is this what you mean?
W (work) = e * V
KE = e * V
(1000 * 1.6 * 10-19) = 1.6 * 10-19 * V
V = 1000 V

 

[Hamal_Arietis]

KE = e * V.
It wrong KE=q.V with q is the charge of α

 

[moenste]

KE = e * V.
It wrong KE=q.V with q is the charge of α

So we have:
KE = 1000 * 1.6 * 10^-19 J (or 1000e)
q = 2 * e = 2 * 1.6 * 10^-19 C (or 2e)
V = ?

1000 * 1.6 * 10^-19 = 2 * 1.6 * 10^-19 * V
or
1000e = 2e * V
V = 1000 / 2 = 500 V

Thank you!