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

Answer:

v = 22492 ms^-1

Explanation:

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

$v = \sqrt{2kqQ / rm}$

$v = \sqrt{2*(9*10^9)*(1.6*10^-19)*(1.6 * 10^-15) / (0.01)*(9.109*10^-31)}$

v = 22492 ms^-1

v = 22.492 k ms^-1