Electric Force/ Electrical Fields

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Homework Help Overview

The discussion revolves around calculating the number of electrons to be removed from two copper spheres to equate the electric force of repulsion between them with the gravitational force of attraction. The subject area includes concepts from electrostatics and gravitation.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between electric and gravitational forces, referencing relevant equations and constants. There are attempts to simplify the problem by equating the two forces and considering the implications of removing electrons.

Discussion Status

Several participants have offered guidance on how to set up the equations and relate the forces. There is ongoing exploration of calculations and interpretations of results, with some questioning the accuracy of previous calculations and assumptions about mass loss.

Contextual Notes

Participants note the potential neglect of the mass change due to electron removal and the importance of ensuring consistent units throughout the calculations. There is also mention of the need to clarify the variables involved in the equations.

xswtxoj
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Homework Statement



How many electrons must be removed from each two of 5.81kg copper spheres to make the electric force of repulsion between them equal to the magnitude to the gravitational attraction between them

Homework Equations



F: Ke (q1)(q2)/r^2

The Attempt at a Solution


( 8.99x10^9)( 9.11x10^-31)(1.67x 10^-27)/ 2 (5.81)
= 3.97x 10^ -47
 
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Maybe you should write out the variables for the equations and simplify before throwing the numbers in?

Besides I'm not familiar with what a 10-47 of an electron looks like.
 
i'm not sure wat to do? probably F=ma then F=Ke(q1)(q2)/r sqed
 
Gravitational force has its own expression,
http://www.physicsclassroom.com/Class/circles/u6l3c.cfm
If you're not familiar with it already.

And I suppose for the question that you neglect the mass being removed when you remove electrons? It's a very small mass, but still a mass.

Anyway, the way I understand the question is that the same amount of electrons is being removed from the spheres, which means one thing, which simplifies everything; the spheres will have the same charge.

You've got the expression for gravitational force and electric force. Relate them in the way the question asks you to.
 
(m)(a)=F then G=(m1)(m2)/d sqed? or F: Ke (q1)(q2)/r^2 or how would i use these?
 
As Hannisch suggests consider the gravitational force of attraction between two 5.8 kg masses.

That must equal the force of repulsion at the same distance - at whatever r because they are both proportional to 1/r2.

Set them equal and discard the terms that cancel. Then solve.
 
so do i start with F= 6.673x 10^-11 * 5.81/ d sq, but d sq cancels, but then (5.81) a= F, therefore 3.87x 10^-10 = F= (5.81) a, ?
 
No.

GMm/r2 = kqq/r2

The masses are equal. The charges equal.

q = m*(G/k)1/2
 
therefore is i solve for q, then (5.81)(( 6.673x 10^11/8.99x10^9)) ^1/2
 
  • #10
so that would be 1.74 electrons?
 
  • #11
I'm not sure how you got something bigger than 1 times 5 to equal 1.74.
 
  • #12
wait if i plug in the equation u gave me then i get 2.15 x 10^-20
 
  • #13
xswtxoj said:
wait if i plug in the equation u gave me then i get 2.15 x 10^-20

I haven't the slightest idea what you calculated.

But I am fairly certain you didn't calculate the square root correctly.

As an example (4*10-6)1/2 will yield 2*10-3
 
  • #14
(5.81)(( 6.673x 10^11/8.99x10^9)) ^1/2 ----------> 6.673x 10^11/8.99x10^9)) ^1/2=8.61x 10^-11 then times 5.81 equals 5x 10^-10
 
  • #15
xswtxoj said:
(5.81)(( 6.673x 10^11/8.99x10^9)) ^1/2 ----------> 6.673x 10^11/8.99x10^9)) ^1/2=8.61x 10^-11 then times 5.81 equals 5x 10^-10

G is 10-11 but you got the right number.

But what are your units?
 

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