Solving for Two Point Charges Using Electric Field Lines and Homework Equations

In summary, the problem involves finding the values of two point charges that are 0.600m apart and experience a repulsive force of 0.400N, with the sum of the two charges equaling 1.00 x 10^-5 C. The equation for Coulomb's Law is given as F = k(q)(q')/r^2, but there are two unknowns (q and q'), so two equations are needed to solve the problem. The second equation is q + q' = 10^-5 C.
  • #1
Rastamonstah
4
0

Homework Statement



Two point charges that are 0.600m apart experience a repulsive force of 0.400N. Sum of the two charges equal 1.00 x 10^-5 C. Find the values of the two charges

Homework Equations





The Attempt at a Solution



My only reference on how to start on this problem is the section of my textbook on Electric Field lines, the textbook outlines the steps on how to get the strength of a field from its field lines, the formula given is completely unrelated to this problem given so i don't know how to start. I don't need an answer i'd just like some help on starting on it.
 
Physics news on Phys.org
  • #2
Rastamonstah said:

Homework Statement



Two point charges that are 0.600m apart experience a repulsive force of 0.400N. Sum of the two charges equal 1.00 x 10^-5 C. Find the values of the two charges

Homework Equations





The Attempt at a Solution



My only reference on how to start on this problem is the section of my textbook on Electric Field lines, the textbook outlines the steps on how to get the strength of a field from its field lines, the formula given is completely unrelated to this problem given so i don't know how to start. I don't need an answer i'd just like some help on starting on it.

Surely your text must supply the basic formula for the force between two charges (Coulomb's Law)?
 
  • #3
gneill said:
Surely your text must supply the basic formula for the force between two charges (Coulomb's Law)?


Yes F= k (q)(q')/r^2


Would i rearrange this equation to find the solution?
 
  • #4
Rastamonstah said:
Yes F= k (q)(q')/r^2


Would i rearrange this equation to find the solution?

Yes, but you have two unknowns: q and q'. So what does that tell you?
 
  • #5
gneill said:
Yes, but you have two unknowns: q and q'. So what does that tell you?

Does that mean i need to set one variable to zero and solve for the other?
 
  • #6
No then you would be cheating nature. When u have an equation with 2 unkowns, u need 2 equations to solve. If u have 3 unkowns, u need 3 equations etc...
 
  • #7
TheTank said:
No then you would be cheating nature. When u have an equation with 2 unkowns, u need 2 equations to solve. If u have 3 unkowns, u need 3 equations etc...

Okay that makes sense, how would i go about setting up these two equations? Am i using Columbs Law twice? Sorry for my ignorance but i haven't been able to grasp much from my lecture.
 
  • #8
No, the second equation is q + q' = 10^-5 C..
 

1. What is an electric field?

An electric field is a region around a charged object where the force of electricity can influence other objects. It is created by the presence of a charged particle and can be either positive or negative.

2. How is electric field strength measured?

Electric field strength is measured in newtons per coulomb (N/C). This measurement represents the amount of force that would be exerted on a charged particle placed in the field.

3. How does the direction of an electric field relate to the direction of the electric force?

The direction of the electric field is always in the direction of the electric force that would be exerted on a positive test charge. This means that the direction of the electric field points away from positive charges and towards negative charges.

4. What factors affect the strength of an electric field?

The strength of an electric field is affected by the magnitude of the charge creating the field, the distance from the charge, and the medium (such as air or water) in which the field exists.

5. How can electric fields be used in everyday life?

Electric fields have a wide range of practical applications, including powering electrical devices, controlling the motion of charged particles in electronic devices, and generating electricity through generators and transformers. They are also used in medical technology, such as MRI machines, and in everyday objects like touch screens and static electricity in clothing.

Similar threads

Electric Field of two charges
    • Introductory Physics Homework Help
Replies
28
Views
576
Discontinuity in an Electric line of force
    • Introductory Physics Homework Help
Replies
12
Views
1K
How do we know the point of 0 electric field is on the axis?
    • Introductory Physics Homework Help
Replies
3
Views
215
Electric field external to Conducting Hollow Sphere with charge inside
    • Introductory Physics Homework Help
Replies
17
Views
401
Why is electric field at the center of a charged disk not zero?
    • Introductory Physics Homework Help
Replies
4
Views
2K
Find the electric field of a long line charge at a radial distance
    • Introductory Physics Homework Help
Replies
1
Views
787
Understanding the energy of a dipole in a uniform electric field
    • Introductory Physics Homework Help
Replies
2
Views
1K
Why Does the Electric Field Sum Instead of Cancel with Opposite Charges?
    • Introductory Physics Homework Help
Replies
6
Views
967
Finding the position where the electric field is zero
    • Introductory Physics Homework Help
Replies
2
Views
535
Electrostatics help please -- Electric field, potential
    • Introductory Physics Homework Help
Replies
5
Views
1K

Hot Threads

Recent Insights

Back
Top