Question

Find electric field intensity at a distance 2A° . From gold nucleus having z=79

Answer 1

Given:

The distance is d= 2A= 2×10⁻¹⁰

The number of charge is n=79

To find:

The electric field intensity

Solution

The charge will be

q=ne

Put the above values

q=79×1.6×10⁻¹⁹

Now the electric field intensity formula

E=q/4πε₀d²

E=79×1.6×10⁻¹⁹×9×10⁹/(2×10⁻¹⁰)²

E=284×10¹⁰

E=2.84×10⁸

The electric field intensity be 2.84×10⁸