Electric Potential — Question 7

Question 7

Opposite charges with equal initial energies emerge at different speeds

Particles of equal mass m and charges $+q$ and $-q$, with $q>0$, each start with $K_0=2q\Delta V$. They traverse ideal workless guides to a point whose potential is higher by $\Delta V>0$. The potential rises monotonically along each guide.

1. Find both final kinetic energies.

2. Find their speed ratio and the condition on a general initial energy for the positive particle to reach the final potential.

Question 7 - Solution

Opposite charges with equal initial energies emerge at different speeds

(a) Method and calculation

$$K_{f,+}=K_0-q\Delta V=q\Delta V.$$$$K_{f,-}=K_0+q\Delta V=3q\Delta V.$$

(b) Interpretation and further calculation

$$\frac{v_{f,-}}{v_{f,+}}=\sqrt{\frac{K_{f,-}}{K_{f,+}}}=\sqrt3.$$$$K_0\ge q\Delta V$$is the energetic reachability condition; equality gives zero terminal speed and may require infinite time for a smooth limiting barrier.

Physical check

Monotonic potential excludes a larger intermediate barrier for the positive particle.