Magnitude of Electric Field from an Atom

Hypnos_16

1. 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

2. Homework Equations

E = kq / r2

3. 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 in to play. If there are 93 Electrons and 93 Protons wouldn't it mean the charge would just be 0?

Any help?

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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

Staff Emeritus
Homework Helper
Gold Member
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

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Homework Helper
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