Magnitude of Electric Field from an Atom

[Hypnos_16]

Homework Statement

The neutral neptunium atom has 93 electrons.
What is the magnitude of its electric field at a distance of 6.5x10-10 m from the nucleus?

Hint: The number of protons in a nucleus is equal to the number of electrons in the neutral atom.

q = 1.6e-19 * 93 = 1.48e-17
r = 6.5e-10
k = 8.99e9

Homework Equations

E = kq / r²

The Attempt at a Solution

I would think that it would just be E = kq / r² since we have all the variables we need.

E = kq / r²
E = (8.99e9)(1.48e-17) / (6.5e-10)²
E = 1.33e-7 / (6.5e-10)²
E = 3.16e11 => 3.16 x 10¹¹ N/C

However turns out that isn't the right answer. So i took the hint into consideration, but can't see how it fits into play. If there are 93 Electrons and 93 Protons wouldn't it mean the charge would just be 0?

Any help?

 

[DukeLuke]

Can you apply Gauss's law at this radius?

 

[Hypnos_16]

Sure, but what will that do?

Gauss's Law = EA
= (3.16e11)(4π(6.5e-10)²)
= 1.67e-6

 

[SammyS]

What is the size of such a atom?

 

[DukeLuke]

Well, you can relate the charge enclosed within a surface to the electric field at the surface regardless of how the charge is distributed. So at the this radius what charge is enclosed?

 

[Hypnos_16]

Well, you can relate the charge enclosed within a surface to the electric field at the surface regardless of how the charge is distributed. So at the this radius what charge is enclosed?

-1.67e-6 = [k(q) / r²] * [4πr²]
-1.67e-6 / [4πr²] = [k(q) / r²]
-1.67e-6[r²] / [4πr²] = k(q)
-1.67e-6 / 4π = k(q)
-1.33e-7 = k(q)
q = -1.33e-7 / 8.99e9
q = -1.48e-17 N/C

 

[WatermelonPig]

http://www.chemicool.com/elements/neptunium.html#radius

The atomic radius of neptunium is 175pm.

 

[SammyS]

http://www.chemicool.com/elements/neptunium.html#radius

The atomic radius of neptunium is 175pm.

Thank you, WatermelonPig !

So, is 6.5x10^-10 m well outside of the radius of the atom?