Electric Force/ Electrical Fields

In summary, to solve for the charge on 5.81kg copper spheres, 8.99x10^9 electrons need to be removed.
  • #1
xswtxoj
87
0

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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  • #2
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.
 
  • #3
i'm not sure wat to do? probably F=ma then F=Ke(q1)(q2)/r sqed
 
  • #4
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.
 
  • #5
(m)(a)=F then G=(m1)(m2)/d sqed? or F: Ke (q1)(q2)/r^2 or how would i use these?
 
  • #6
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.
 
  • #7
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, ?
 
  • #8
No.

GMm/r2 = kqq/r2

The masses are equal. The charges equal.

q = m*(G/k)1/2
 
  • #9
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?
 

What is electric force?

Electric force is the attractive or repulsive force between two electrically charged objects. This force is caused by the interaction between the electric fields of the two objects and can be described by Coulomb's law.

How is electric force different from gravitational force?

Electric force is caused by the interaction of electric charges, while gravitational force is caused by the interaction of masses. Additionally, electric force can be either attractive or repulsive, while gravitational force is always attractive.

What is an electric field?

An electric field is a region of space around an electrically charged object where the electric force can be felt. It is represented by electric field lines and the strength of the field is determined by the magnitude and direction of the electric force.

How is the strength of an electric field measured?

The strength of an electric field is measured by the electric field intensity, which is defined as the force per unit charge at a given point in the field. It is represented by the symbol E and its unit is Newtons per Coulomb (N/C).

How does distance affect electric force?

According to Coulomb's law, the electric force between two charged objects is inversely proportional to the square of the distance between them. This means that as the distance between two charged objects increases, the electric force between them decreases.

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