An electron is released from the rest at one point in a uniform electric field and moves a distance of 10 cm in 10^-1 s. What is the voltage between the two points?

Explanation:

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. 🎉 ```