Another Explanation (5): ```html
Step-by-step Solution:
Given,
- Distance, \(d = 10\) cm = 0.1 m
- Time, \(t = 10^{-7}\) s
- Electron's charge, \(e = 1.6 \times 10^{-19}\) C
- Electron's mass, \(m = 9.1 \times 10^{-31}\) kg
Since the electron starts from rest, initial velocity \(u = 0\).
Using the equation of motion, \(s = ut + \frac{1}{2}at^2\), where \(s\) is the distance, \(u\) is the initial velocity, \(t\) is the time, and \(a\) is the acceleration.
Here, \(0.1 = 0 + \frac{1}{2} a (10^{-7})^2\)
So, acceleration \(a = \frac{2 \times 0.1}{(10^{-7})^2} = 2 \times 10^{13} \) m/s\(^2\)
The force on the electron due to the electric field is \(F = eE\), where \(E\) is the electric field.
Also, from Newton's second law, \(F = ma\).
Therefore, \(eE = ma\)
\(E = \frac{ma}{e} = \frac{9.1 \times 10^{-31} \times 2 \times 10^{13}}{1.6 \times 10^{-19}} = \frac{18.2 \times 10^{-18}}{1.6 \times 10^{-19}} = 113.75\) N/C
The voltage \(V\) between the two points is given by \(V = Ed\), where \(d\) is the distance.
\(V = 113.75 \times 0.1 = 11.375\) V
So, the voltage between the two points is 11.375 mV.
\(V = 11.375\) V = \(11.375 \times 10^{-3}\) V = 11.375 mV
Therefore, the voltage between the two points is 11.375 mV. 🎉
```
