Question

A 2.6 mm -diameter sphere is charged to -4.5 nC . An electron fired directly at the sphere from far away comes to within 0.37 mm of the surface of the target before being reflected. Part A What was the electron's initial speed

Answer 1

Answer: $1.96\times 10^{8}\ m/s$

Explanation:

Given

Diameter of sphere is $d=2.6\ mm\quad \quad[r=1.3 mm]$

Charge on the sphere is $Q=-4.5\ nC$

Nearest distance electron can reach to sphere is $d=0.37\ mm$

Here, kinetic energy of electron is converted into electrostatic energy between the two i.e.

$\Rightarrow \dfrac{1}{2}mv^2=\dfrac{kQq}{d}\\\\\Rightarrow \dfrac{1}{2}\times 9.1\times 10^{-31}\times v^2=\dfrac{9\times 10^9\times 4.5\times 10^{-9}\times 1.6\times 10^{-19}}{(3.7\times 10^{-4})}\\\\\Rightarrow v^2=3.8491\times 10^{16}\\\Rightarrow v=1.96\times 10^{8}\ m/s$

Thus, the initial speed of electron is $1.96\times 10^{8}\ m/s$.