Answer:
[tex]E=8*10^5\frac{V}{m}[/tex]
Explanation:
The magnitude of the electric field between two parallel conducting plates is defined as:
[tex]E=\frac{\Delta V}{d}[/tex]
Here [tex]\Delta V[/tex] is the potential difference between the plates and d its separation.
The electric potential energy is defined as the product between the particle's charge and the potential difference:
[tex]U=q\Delta V[/tex]
Solving for [tex]\Delta V[/tex] and replacing in the electric field formula:
[tex]\Delta V=\frac{U}{q}\\E=\frac{U}{qd}[/tex]
In this case we have a double charged ion, so [tex]q=2e[/tex]:
[tex]E=\frac{32*10^3eV}{(2e)(2*10^{-2}m)}\\E=8*10^5\frac{V}{m}[/tex]
