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Electric Field of Earth problem

by stylez03
Tags: earth, electric, field
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stylez03
#1
Jan26-07, 07:04 PM
P: 139
1. The problem statement, all variables and given/known data
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?


2. Relevant equations

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

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

3. 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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Chi Meson
#2
Jan26-07, 07:39 PM
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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?
stylez03
#3
Jan26-07, 07:42 PM
P: 139
Electric force on a charged body is exherted by the electric field created by other charged bodies?

Chi Meson
#4
Jan26-07, 07:59 PM
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Electric Field of Earth problem

I meant what is the (very simple) equation relating the three quantities?
stylez03
#5
Jan26-07, 08:02 PM
P: 139
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?
Chi Meson
#6
Jan26-07, 08:08 PM
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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.
stylez03
#7
Jan26-07, 08:11 PM
P: 139
So it should be q = F/E because for part 1 it's looking for the charge.

q = 58.0 kg / 153 ??
marcusl
#8
Jan26-07, 08:39 PM
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Quote Quote by Chi Meson View Post
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
Chi Meson
#9
Jan27-07, 11:33 AM
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Quote Quote by stylez03 View Post
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?
stylez03
#10
Jan27-07, 11:35 AM
P: 139
Mass is a measurement of the amount of matter something where as weight is the measurement of the pull of gravity on an object.
stylez03
#11
Jan28-07, 02:15 PM
P: 139
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?
stylez03
#12
Jan28-07, 11:35 PM
P: 139
[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.
mace2
#13
Jan29-07, 03:10 AM
P: 98
where did the 10^-9 come from?
stylez03
#14
Jan29-07, 10:32 AM
P: 139
Quote Quote by mace2 View Post
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?
mace2
#15
Jan29-07, 09:01 PM
P: 98
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.
stylez03
#16
Jan29-07, 09:44 PM
P: 139
Quote Quote by mace2 View Post
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
mace2
#17
Jan30-07, 12:04 AM
P: 98
um, maybe try (-3.72)^2
stylez03
#18
Jan30-07, 11:45 AM
P: 139
Quote Quote by mace2 View Post
um, maybe try (-3.72)^2
nope, didn't work


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