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?
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??