# Electric field of lead atom

mimictt

## Homework Statement

The nucleus of a lead atom has a charge of 82 protons
a. what are the direction and magnitude of the electric field at 1.0x10^-10 m from the nucleus?

E=Kq/r^2

## The Attempt at a Solution

I did E= ((9 x10^9)(1.6X10^-19)^82)/ (1.0x10^-10)^2

but when I checked cramster.com for the answer, it gave me E= ((9 x10^9) (82) (1.6X10^-19))/ (1.0x10^-10)^2

can someone explain to me why it's 82 times (1.6X10^-19) not (1.6X10^-19)^82??

thank you!

## Answers and Replies

Each proton has a charge of 1.6E-19 coulombs, so 82 of them have a charge of 82*1.6E-19 coulombs. Why do you think it should be (1.6E-19)^82?

mimictt
Each proton has a charge of 1.6E-19 coulombs, so 82 of them have a charge of 82*1.6E-19 coulombs. Why do you think it should be (1.6E-19)^82?

well because I thought that if you have to calculate the force of two protons/electrons then you would have to multiply Q1 x Q2 and since they both have the same charge---> Q^2

this is how i got (1.6E-19)^82 because i thought that there are 82 protons with the same charge...

cupid.callin
No!! you multiply the total charges in the two bodies.