Electric Forcs and Work (Algebra based physics)

In summary, the problem involves two charges, q1=+7.15 x 10-05 C and q2=-7.80 x 10-04 C, at rest 71.5 cm apart. The question asks for the amount of work required to move the charges slowly and steadily until they are 31.0 cm apart. The solution involves calculating the initial force using the formula F = kq/r and using the distance one charge moved to calculate the final force. The potential energy between the two charges is then used to calculate the work done, which is found to be 300J.
  • #1
Seikon85
2
0
Charges q1=+7.15 x 10-05 C and q2=-7.80 x 10-04 C are placed at rest 71.5 cm apart. How much work must be done by an outside agent to move these charges slowly and steadily until they are 31.0 cm apart?

Ok so I tried a few things on my own, and got stuck. So I referred to a previous post from this morning for guidance:

http://answerboard.cramster.com/physics-topic-5-199660-0.aspx

however, I am still getting the wrong answer.


I am using F = kq/r initially. The r value i am using was obtained by figuring out the distance one charge moved. (.715-.31)/2. Then subtracting that distance from .715. So my r = .513m here.

I get F = 968J.

W = F*d = 968J * (.31m) = 300J.


Am I missing something obvious. Any insight?

Thanks.
 
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  • #2
Have you learned about potential energy between two charges?
 
  • #3


First of all, great job on attempting to solve the problem on your own and seeking guidance when you got stuck. It's always important to utilize all available resources in order to fully understand a concept.
Now, let's take a look at the problem. You are correct in using the formula F = kq/r initially. However, the distance between the charges is not simply the difference between their initial and final positions. In this case, the distance between the charges changes as they are moved closer together. Therefore, you need to use the formula for electric potential energy, which is U = kq1q2/r. This takes into account the changing distance between the charges.
Now, to find the work done by an outside agent, we can use the formula W = Uf - Ui, where Uf is the final potential energy and Ui is the initial potential energy.
Using this formula, we get:
W = (kq1q2)/rf - (kq1q2)/ri
= (9x10^9)(+7.15x10^-5)(-7.80x10^-4)/(.31) - (9x10^9)(+7.15x10^-5)(-7.80x10^-4)/(.715)
= -1.98 J
Therefore, the work done by an outside agent to move the charges slowly and steadily until they are 31.0 cm apart is -1.98 J.
I hope this helps clarify the problem for you. Keep up the good work!
 

Related to Electric Forcs and Work (Algebra based physics)

1. What is an electric force?

An electric force is a type of force that exists between two charged objects. It can either be attractive or repulsive, depending on the charges of the objects. Like charges repel each other, while opposite charges attract each other.

2. What is the formula for electric force?

The formula for electric force is F = k * (q1 * q2) / r^2, where F is the force, k is the Coulomb's constant, q1 and q2 are the charges of the two objects, and r is the distance between them.

3. How does distance affect electric force?

Distance has an inverse square relationship with electric force. This means that as the distance between two charged objects increases, the electric force between them decreases. This relationship is described by the formula F ∝ 1/r^2.

4. What is electric work?

Electric work is the amount of energy required to move a charged object from one point to another in an electric field. It is calculated by multiplying the electric force by the distance moved in the direction of the force.

5. What is the unit of electric force and work?

The unit of electric force is Newton (N), while the unit of electric work is Joule (J).

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