Coulomb's Law and electric fields

In summary, three point charges (+2 uC, -3 uC, -5 uC) placed on the x-axis create a force of -0.55 N on the -3 uC charge and a force of 0.15 N on the -5 uC charge. The force is calculated using the equation Fe=kq1q2/r^2 and by summing the forces created by each charge. The charge units used in this problem are micro-coulombs (uC). A free-body diagram can be helpful in solving this problem.
  • #1
poohead
34
0

Homework Statement



three point charges are placed at the following points on the x-axis +2 uC at x=0, -3uC at x=40 cm, -5 uC at x=120 cm. find the force (a) on the -3 uC charge, (b) on the -5 uC charge

ans. (a) -0.55N ; (b) 0.15 N


uC=nano coulomb


Homework Equations



Fm=qvB ?

Fe=|E|(q)?

Fe=kq1q2/r^2

The Attempt at a Solution



I cannot seem to solve such a simple problem, i need sufficient help. I have the idea that it has something to do with minusing each Fe charge by each other but I am uncertain please need help with a solution
 
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  • #2
poohead said:
I cannot seem to solve such a simple problem, i need sufficient help. I have the idea that it has something to do with minusing each Fe charge by each other but I am uncertain please need help with a solution

You'll use the 3rd equation, and you will sum the forces. For instance, for the -3 nC charge, you will have to figure out the force created by the +2 nC Charge, then for the -5 nC charge. Once you get those two numbers, just add them together.
 
  • #3
but wait, try it for yourself, apparently you don't get the given answer of -0.55 N, what is the answer u thus get?
 
  • #4
Well, I didn't get -0.55 N exactly, but I did get -0.5479 N. Which is close enough. I came to this solution saying the uC is actually micro-coulombs. Usually nano-coulombs is denoted nC, and if you don't want to go into LaTex, microcoulombs is uC. This is because u is close enough to the Greek Letter mu.

A few more hints on this problem... Draw a free-body diagram. Then think about what "should" be happening, like which direction a force should go.
 
  • #5
got it
now I am trying part b
thanks man
 

Related to Coulomb's Law and electric fields

What is Coulomb's Law and how does it relate to electric fields?

Coulomb's Law is a fundamental law of physics that describes the force of attraction or repulsion between two electrically charged particles. It states that the force is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them. Electric fields are a representation of the force that a charged particle would experience in a given space. Coulomb's Law helps us understand the relationship between electric charges and electric fields.

How do you calculate the force between two charged particles using Coulomb's Law?

To calculate the force between two charged particles using Coulomb's Law, you need to know the magnitude of the charges (q1 and q2), the distance between them (r), and the constant of proportionality (k). The equation is F = k * (q1 * q2)/r^2, where k is equal to 9 x 10^9 N*m^2/C^2.

What is an electric field and how is it represented?

An electric field is a region in space where an electrically charged particle experiences a force. It is represented by lines called electric field lines, which indicate the direction and strength of the electric field. The direction of the electric field lines is from positive to negative, and the closer the lines are together, the stronger the electric field is.

How does the direction of electric field lines relate to the direction of the electric field?

The direction of the electric field lines is always in the direction that a positive test charge would move if placed in the electric field. This means that the electric field lines point away from positive charges and towards negative charges. The direction of the electric field itself is the direction in which a positive test charge would experience a force if placed in the electric field.

What is the relationship between electric field strength and electric potential?

Electric field strength and electric potential are related by the equation E = -dV/dr, where E is the electric field strength and V is the electric potential. This means that the electric field strength is equal to the negative gradient of the electric potential. In other words, the electric field strength is the rate of change of the electric potential with respect to distance. The greater the change in electric potential over a given distance, the stronger the electric field will be.

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