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Electric field of lead atom

  • Thread starter mimictt
  • Start date
  • #1
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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?

Homework Equations


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

  • #2
phyzguy
Science Advisor
4,502
1,446
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?
 
  • #3
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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...
 
  • #4
1,137
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No!! you multiply the total charges in the two bodies.
 

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