## College Physics (4th Edition)

We can write an expression for the original energy: $E_1 = q~\Delta V_1$ We can write an expression for the energy when the potential difference is doubled: $E_2 = q~\Delta V_2$ $E_2 = q~(2\times \Delta V_1)$ $E_2 = 2~q~\Delta V_1$ $E_2 = 2~E_1$ When the potential difference is doubled, the energies of the characteristic x-rays are doubled.