Electric Field of Earth problem

In summary: Please check your calculations and try again.In summary, the Earth has a net electric charge that causes a field at points near its surface equal to 153 N/C and directed in toward the center of the earth. To overcome one's weight by the force exerted by the Earth's electric field, a human with a mass of 58.0 kg would need to acquire a charge of 3.72 x 10^-9 C. The magnitude of the repulsive force between two people with this same charge and separated by a distance of 120 m can be found using the equation F=q^2/4πε0r^2, which results in a force of approximately 5.41 x 10^-9 N. However
  • #1
stylez03
139
0

Homework Statement


Electric Field of the Earth. The Earth has a net electric charge that causes a field at points near its surface equal to 153 N/C and directed in toward the center of the earth.

1. What charge would a human with a mass of 58.0 kg have to acquire to overcome his or her weight by the force exerted by the Earth's electric field?

2. What would be the magnitude of the repulsive force between two people each with the charge calculated in part (a) and separated by a distance of 120 m?

Homework Equations



[tex] E = \frac{1}{4*pi*e_o}*\frac{q}{r^2*\vec{r}}[/tex]

[tex] G = 9.81 & m/s [/tex]

The Attempt at a Solution



I couldn't find any relevant information in the electric field section that I could map to this problem to help me solve it. Does anyone have a suggestion on where to start?
 
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  • #2
Your looking at the wrong equation here. This is a simple problem. Just a matter of balancing two forces.

How are charge, electric field, and electric force related?
 
  • #3
Electric force on a charged body is exherted by the electric field created by other charged bodies?
 
  • #4
I meant what is the (very simple) equation relating the three quantities?
 
  • #5
I can't seem to find the equation that relates to all three, I'm looking at the index of current chapter of all the equations in this chapter and I can't find one that does?
 
  • #6
I'm surprised. Well it's easy: F=qE . Notice how it is similar to the equation for weight: F=mg . In both cases the force is found as the product of a quantity of matter times the strength of a field.

Balance those forces.
 
  • #7
So it should be q = F/E because for part 1 it's looking for the charge.

q = 58.0 kg / 153 ??
 
  • #8
Chi Meson said:
I'm surprised. Well it's easy: F=qE . Notice how it is similar to the equation for weight: F=mg .
Hmm, I gave stylez the same equation in another thread about 12 hours earlier :uhh:
 
  • #9
stylez03 said:
So it should be q = F/E because for part 1 it's looking for the charge.

q = 58.0 kg / 153 ??

What's the difference betwen mass and weight?
 
  • #10
Mass is a measurement of the amount of matter something where as weight is the measurement of the pull of gravity on an object.
 
  • #11
EDIT:

Thanks I found the right equation. For part B it asks:

What would be the magnitude of the repulsive force between two people each with the charge calculated in part (a) and separated by a distance of 120 m?

Is there a separate to find the repulsive force?
 
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  • #12
[tex] \frac{1} {4*pi*8.85*10^-12} * \frac{-3.72*10^-9^2} {120^2} [/tex]

Though the online program says I'm off by an additive constant.
 
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  • #13
where did the 10^-9 come from?
 
  • #14
mace2 said:
where did the 10^-9 come from?

For some of the problems when they mention charge, the example I had to build off of was to take it and multiply it by 10^-9. The equation I'm using is the following:

[tex] \frac{1} {4*pi*8.85*10^-12} * \frac{q1*q2}{r^2} [/tex]

Should q1 and q2 just be -3.72 and since q1 = q2 it should just be q^2 ? Is this the equation you would use to find the repulsive force?
 
  • #15
Yeah, that will give you the electrostatic force. Since they're both negatively charged they will repel.

Don't guess about 10^-9, just work with what you have. The math doesn't lie.

q1 & q2 & so on are all measured in Coloumbs.
 
  • #16
mace2 said:
Yeah, that will give you the electrostatic force. Since they're both negatively charged they will repel.

Don't guess about 10^-9, just work with what you have. The math doesn't lie.

q1 & q2 & so on are all measured in Coloumbs.

I tried:

[tex] \frac{1} {4*pi*8.85*10^-12} * \frac{-3.72^2}{120^2} [/tex]

Though I still get Your answer is off by an additive constant.

The units of the solution is in Newtons
 
  • #17
um, maybe try (-3.72)^2
 
  • #18
mace2 said:
um, maybe try (-3.72)^2

nope, didn't work
 
  • #19
i have no idea. maybe they don't like your rounding? what is an "additive constant"?
 
  • #20
mace2 said:
i have no idea. maybe they don't like your rounding? what is an "additive constant"?

I don't know, but I ran out of trys for the online program.
 
  • #21
What answer did you get through your calculations? You probably made a small error while calculating the result.
 

What is the "Electric Field of Earth problem"?

The "Electric Field of Earth problem" refers to the study of the electric field that surrounds the Earth. It is caused by the Earth's rotation and its interaction with the solar wind and other cosmic particles.

How is the electric field of Earth measured?

The electric field of Earth is measured using instruments called magnetometers and electric field detectors. These instruments are usually placed on satellites or spacecraft orbiting the Earth.

What are the effects of the electric field of Earth?

The electric field of Earth has several effects, including the formation of the auroras (northern and southern lights), the creation of the Van Allen radiation belts, and the disruption of satellite communications.

Can the electric field of Earth be shielded or neutralized?

Yes, the electric field of Earth can be shielded or neutralized by using materials that conduct electricity, such as metals. This is often done in spacecraft and satellites to protect them from the effects of the electric field.

How does the electric field of Earth affect living organisms?

The electric field of Earth has a negligible effect on living organisms, as our bodies are mostly made up of water which is a poor conductor of electricity. However, some animals, such as birds and bees, may use the Earth's electric field for navigation.

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