Question

What is the escape speed for an electron initially at rest on the surface of a sphere with a radius of 1.0 cm and a uniformly distributed charge of 1.91 ×10-15 C

Answer 1

v = 22492 ms^-1

Step-by-step explanation:

`1/2 m v^2 = kqQ / r`

`v = √(2kqQ / rm)`

`v = √(2*(9*10^9)*(1.6*10^-19)*(1.6 * 10^-15) / (0.01)*(9.109*10^-31))`

v = 22492 ms^-1

v = 22.492 k ms^-1