Finding the Net Charge of the Earth

Hypnos_16

1. 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

2. Homework Equations

Gauss's Law = EA

3. 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?

Related Introductory Physics Homework Help News on Phys.org

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 = - * [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

Last edited:

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 = qencl. / 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 alot since Im 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.

Last edited:

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving