Electric Potential — Question 8

Question 8

A negative ion is stopped by a drop in potential

An ion with charge $-2e$ has initial kinetic energy 400 eV. It enters a retarding region and reaches a turning point B from its initial point A. Only the electric force does work.

1. Find $V_B-V_A$, retaining its sign.

2. Find its remaining kinetic energy at a point 150 V below A and explain why a 200 V rise would not stop it.

Question 8 - Solution

A negative ion is stopped by a drop in potential

(a) Method and calculation

$$0-K_A=-q(V_B-V_A)\Rightarrow\Delta V=K_A/q.$$$$\Delta V=400\,\mathrm{eV}/(-2e)=-200\,\mathrm V.$$

(b) Interpretation and further calculation

$$K(-150\,\mathrm V)=400\,\mathrm{eV}-(-2e)(-150\,\mathrm V)=100\,\mathrm{eV}.$$$$K(+200\,\mathrm V)=400\,\mathrm{eV}+400\,\mathrm{eV}=800\,\mathrm{eV}.$$

Physical check

A negative charge gains electric energy while moving to lower potential.