Source: High school physics (Chinese)
Problem Sets:
Problem
In a vacuum, there is a point charge $Q = -1.2 \times 10^{-6}$ C. An electron is initially located 3 cm away from Q, and then moves to a position 4 cm away from Q.
- What is its initial potential energy in electron-volts?
- What is its final potential energy in electron-volts?
[Q1] $E_{p1} = 3.6 \times 10^5$ eV [Q2] $E_{p2} = 2.7 \times 10^5$ eV
The potential energy of an electron (charge $q=-e$) in an electric potential $U$ is $E_p(\text{J}) = -eU$. To express this in electron-volts (eV), we divide by the elementary charge $e$: $E_p(\text{eV}) = \frac{-eU}{e} = -U$. The electric potential $U$ at a distance $r$ from a point charge $Q$ is $U = \frac{kQ}{r}$. [Q1] At the initial position $r_1 = 3$ cm = 0.03 m:
$$U_1 = \frac{kQ}{r_1} = \frac{(9 \times 10^9)(-1.2 \times 10^{-6})}{0.03} = -3.6 \times 10^5 \text{ V}$$ $$E_{p1}(\text{eV}) = -U_1 = -(-3.6 \times 10^5 \text{ V}) = 3.6 \times 10^5 \text{ eV}$$[Q2] At the final position $r_2 = 4$ cm = 0.04 m:
$$U_2 = \frac{kQ}{r_2} = \frac{(9 \times 10^9)(-1.2 \times 10^{-6})}{0.04} = -2.7 \times 10^5 \text{ V}$$ $$E_{p2}(\text{eV}) = -U_2 = -(-2.7 \times 10^5 \text{ V}) = 2.7 \times 10^5 \text{ eV}$$