Question

- The total positive charge on a silver nucleous
( atomic number: Z=47) is uniformly distributed
on a sphere of radius 10÷10000000000m what will be the
the intensity of electric field at the surface of
this sphere?​

Answer 1

Answer:

- The total positive charge on a silver nucleous

( atomic number: Z=47) is uniformly distributed

on a sphere of radius 10÷10000000000m what will be the

the intensity of electric field at the surface of

this sphere?

Answer 2

Given info : a silver nucleus of atomic no 47 and radius of it is 10 A°.

To find : total positive charge on silver nucleus and also intensity of electric field at the surface of this sphere are ..

solution : atomic no = 47, it means there are 47 protons in the nucleus.

we know, charge on each proton = 1.6 × 10^-19 C

so, total charge = 47 × 1.6 × 10^-19 C

= 75.2 × 10^-19C

= 7.52 × 10^-18C

now, electric field at the surface of sphere, E = kq/R²

here, R = 10A° = 10 × 10¯¹⁰ m

q = 7.52 × 10^-18 C

so, E = (9 × 10^9 × 7.52 × 10^-18)/(10 × 10¯¹⁰)²

= 67.68 × 10^-9 × 10^18

= 6.768 × 10¹⁰ N/C

Therefore the electric field at the surface of sphere is 6.768 × 10¹⁰ N/C