Electric Field of Earth problem

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Homework Help Overview

The discussion revolves around calculating the electric field of the Earth and its implications for a human's charge needed to counteract weight. The original poster presents two questions: one regarding the charge required to balance weight against the Earth's electric field, and another concerning the repulsive force between two charged individuals.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between electric force, charge, and electric field, questioning how to balance forces in the context of the problem.
  • Some participants express difficulty in identifying the relevant equations and concepts, while others suggest using the equation F=qE to relate charge and electric force.
  • There is discussion about the distinction between mass and weight, and how that relates to the calculations needed for the problem.
  • Questions arise regarding the specifics of calculating repulsive force and the appropriate use of constants in the equations.

Discussion Status

The conversation is ongoing, with participants providing hints and guidance on relevant equations. There is a mix of interpretations regarding the calculations, and some participants are actively trying to clarify their understanding of the concepts involved.

Contextual Notes

Participants mention challenges in finding the right equations and constants, indicating potential gaps in the provided information or assumptions about the problem setup. There is also a reference to an online program that evaluates their answers, which adds a layer of complexity to their attempts.

stylez03
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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?
 
Last edited:
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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?
 
Electric force on a charged body is exherted by the electric field created by other charged bodies?
 
I meant what is the (very simple) equation relating the three quantities?
 
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?
 
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.
 
So it should be q = F/E because for part 1 it's looking for the charge.

q = 58.0 kg / 153 ??
 
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 :rolleyes:
 
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 between 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?
 
Last edited:
  • #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.
 

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