Question

find electric field intensity at a distance 2A from gold nucleus having Z=79 ?​

Answer 1

The value of electric field intensity be 2.84×10^8

Explanation:

We are given that:

• Distance =  2 A = 2  10^-10
• Charge value Z = 79

Solution:

The Value of charge will be

q  = n e

Put the values

q = 79 × 1.6 × 10^-19

Now the electric field intensity formula

E = q / 4πε₀d^2

E = 79 × 1.6 × 10^-19 × 9 × 10^9 / (2 × 106-10)^2

E = 284 × 10^10

E = 2.84×10^8

Hence the value of electric field intensity be 2.84×10^8