Finding the Net Charge of the Earth

[Hypnos_16]

Homework Statement

The Earth is surrounded by an electric field, pointing inward at every point. Assume a magnitude of E = 142N/C near the surface.
a) What is the net charge on the Earth?
b) How many excess electrons per square meter on the Earth's surface does this correspond to?

E = 142 N/C
Radius of Earth = 6.3781 km = 6,378,100 m

Homework Equations

Gauss's Law = EA

The Attempt at a Solution

So i know that i have to use Gauss's Law here. Which is simple the Electric Field, (given) and the area of the Earth which is found be 4πr²

Gauss's Law = EA
= (142) * (4πr²)
= (142) * (4π(6,378,100)²)
= (142) * (5.11e+14)
= 7.26e+16 C

however. After attempting that, i found it wasn't the right answer. Where did i go wrong?

i haven't attempted the second part since i need to know the first part. But i assume that to do it i would have to find the surface area of the Earth and divide it by the number i found in part a, then divide that by 1.6e-19 to find the number of electrons?

 

[Hypnos_16]

Nevermind. I solved the problem.

 

[WJSwanson]

If you don't mind, please post the solution and what brought you to it. It's always helpful for others who might have a similar question to be able to see how others having the same issue might successfully approach it.

 

[Hypnos_16]

Gauss's Law = EA
= (142) * (4πr²)
= (142) * (4π(6,378,100)²)
= (142) * (5.11e+14)
= 7.26e+16 C

θ = 180° for each ∆A => -1
(Basically means it's negative since electric field and ∆A are in complete opposite directions So Electric Field then is just a negative)

Gauss's Law = q / E₀
q = (-E * A * E₀ )
q = -E * E₀ * Area
q = -[142] * [8.85e-12] * [4π(6,378,100)²]
q = [-7.26e+16 C] * [8.85e-12 C² / Nm²]
q = -6.425e+5

Number of Electrons (Part b)

We know the electric field of the Earth = -6.425e+5
That is -6.425e+5 / -1.6e-19 = 4.02e+24 Electrons
The Earth is 5.11e+14 Square meters

5.11e+14 / 4.02e+24
= 7.86e9 Electrons per Square Meter

 

[WJSwanson]

Looks good. Thank you. For future reference, by the way, you can simply write Gauss's law as flux = closed integral of E dot dA = q_encl. / epsilon₀ (and using the TeX tags you can represent flux with the capital Phi using ' \Phi ' and lowercase epsilon w/ subscript 0 ("epsilon naught") as ' \epsilon_{0} ' -- case-sensitive for both of those, and the closed integral as ' \oint_{S} ' to show it's the closed integral over the surface).

You can also simply rewrite it in ASCII if you're in a hurry as

q_enc = eps_0 * (E . dA)

where the boldface letters represent vector quantities.

One of the great things about PF is that it has a wide range of easy-to-learn tools to let you express your mathematics clearly.

But yeah, you got the answer right. At this point I'm basically just talking style: It's less confusing to the reader when you say "flux" or "Phi" equals this or that instead of saying "Gauss's Law" equals something.

 

[wli32492]

Also is it possible to show how your units cancel out to be Coulombs only for the net charge of the earth? It would help a lot since I am trying to figure out how the units cancel. Thanks in advance.

Also for your last part (5.11e14)/(4.02e24) =/= 7.89e9.
(5.11e14)/(4.02e24) = 1.27e-10.

I think you just switched the numbers around wrong but the answers are correct.