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?

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!

 

[phyzguy]

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.

 

[rwisz]

The reason for the difference in the two solutions is due to the fact that the electric field is a vector quantity, meaning it has both magnitude and direction. In this case, the direction of the electric field at a distance of 1.0x10^-10 m from the nucleus is determined by the charge of the nucleus, which is positive due to the 82 protons. This means that the electric field points away from the nucleus in all directions.

To calculate the magnitude of the electric field, we use the equation E=Kq/r^2. However, since the electric field is pointing away from the nucleus, we need to take into account the direction of the electric field when calculating the magnitude. This is why we use the charge of the nucleus (82) instead of the charge of a single proton (1.6x10^-19) in the equation. The charge of the nucleus represents the total charge of all 82 protons, which gives us the correct magnitude of the electric field.

In summary, the correct solution is E= ((9 x10^9) (82) (1.6X10^-19))/ (1.0x10^-10)^2, taking into account the direction of the electric field.