Find Minimum Potential of Two Charges on x-Axis

  • Thread starter Thread starter eltel2910
  • Start date Start date
  • Tags Tags
    Minimum Potential
Click For Summary
SUMMARY

The discussion focuses on finding the minimum electric potential created by two point charges on the x-axis: a charge of 0.883 nC at the origin and another charge of 0.347 nC located at x1 = 13.1 cm. The necessary condition for the potential to reach a minimum is that the derivative of the potential function must equal zero. The minimum potential occurs between the two charges, where a test charge would experience no net force. The user seeks assistance in setting up the derivative of the potential function, specifically k*q1/r - k*q2/(r-x1).

PREREQUISITES
  • Understanding of electric potential and point charges
  • Knowledge of calculus, specifically derivatives
  • Familiarity with Coulomb's law and the constant k
  • Basic concepts of electrostatics and charge interactions
NEXT STEPS
  • Study the derivation of electric potential from point charges
  • Learn how to apply calculus to find critical points in functions
  • Explore the concept of electric field and its relation to potential
  • Investigate the implications of charge placement on electric potential
USEFUL FOR

Students in physics, electrical engineers, and anyone interested in electrostatics and potential energy calculations.

eltel2910
Messages
9
Reaction score
0
A charge of 0.883 nC is placed at the origin. Another charge of 0.347 nC is placed at x1 = 13.1 cm on the x-axis.

At which point on the x-axis does this potential have a minimum?

I do know a few things:

A necessary condition for the potential to have a minimum is that its derivative is 0.

The minimum is between the two charges.

I need to take the derivative of:
k*q1/r - k*q2/(r-x1), but I'm not sure about how to go about that?

I'm not looking for the answer. Just need some help setting this up.
 
Physics news on Phys.org
There should be a point in between the two charges where a test charge experiences no force at all. Which means that you could just plonk the test charge in there. No need to do any work to get it there, you could just slip it in sideways!
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 23 ·
Replies
23
Views
2K
Replies
13
Views
2K
Replies
64
Views
6K
  • · Replies 2 ·
Replies
2
Views
8K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
1
Views
3K