Question

You cause a particle to move from point A, where the electric potential is 11.3 V, to point B, where the electric potential is −25.9 V. Calculate the change that occurs in the particle's electrostatic potential energy, when the particle is an electron, a proton, a neutral hydrogen atom, and a singly ionized helium atom (i.e., lacking one electron from its neutral state).

Answer 1

Explanation:

The electric potential is the electric potential energy per unit of charge

$V=\frac{U}{q}$

Using this definition, we can calculate the electrostatic potential energy change between point A and B:

$\Delta U=U_A-U_B\\\Delta U=qV_A-qV_B\\\Delta U=q(V_A-V_B)$

Electron: $q=-1.6*10^{-19}C$

$\Delta U=-1.6*10^{-19}C(11.3V+25.9V)\\\Delta U=-5.952*10^{-18}J$

Proton: $q=1.6*10^{-19}C$

$\Delta U=1.6*10^{-19}C(11.3V+25.9V)\\\Delta U=5.952*10^{-18}J$

Neutral hydrogen atom: $q=0$

$\Delta U=0$

Singly ionized helium atom: $q=1.6*10^{-19}C$

$\Delta U=1.6*10^{-19}C(11.3V+25.9V)\\\Delta U=5.952*10^{-18}J$