Question

A double charged ion is accelerated to an energy of 32.0 keV by the electric field between two parallel conducting plates separated by 2.00 cm. What is the electric field strength between the plates?

Answer 1

Answer:

$E=8*10^5\frac{V}{m}$

Explanation:

The magnitude of the electric field between two parallel conducting plates is defined as:

$E=\frac{\Delta V}{d}$

Here $\Delta V$ 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:

$U=q\Delta V$

Solving for $\Delta V$ and replacing in the electric field formula:

$\Delta V=\frac{U}{q}\\E=\frac{U}{qd}$

In this case we have a double charged ion, so $q=2e$:

$E=\frac{32*10^3eV}{(2e)(2*10^{-2}m)}\\E=8*10^5\frac{V}{m}$