Find Potential Minimum of Two Point Charges

In summary, the problem involves two charges, one at the origin and one at a specific point on the x-axis. The question asks for the location on the x-axis where the potential has a minimum. The necessary condition for a minimum is that the derivative of the potential is 0. The potential at a point distance x from the first charge is given by V = k*q1/x + k*q2/(d-x), where d is the distance between the charges. To find the minimum potential, we need to find the derivative of this equation and set it equal to 0. Solving for x will give us the location of the minimum potential.
  • #1
Mikesgto
18
0

Homework Statement


A charge of 0.611 nC is placed at the origin. Another charge of 0.383 nC is placed at x1 = 8.1 cm on the x-axis.
At which point on the x-axis does this potential have a minimum?

Homework Equations


U=(kq1q2)/r

The Attempt at a Solution


I really have no idea how to even start this problem. I've been thinking about it for a couple hours now. Any help would be appreciated.

Homework Statement


Homework Equations


The Attempt at a Solution

 
Last edited:
Physics news on Phys.org
  • #2
Post the complete question.
 
  • #3
I edited the post. My apologies.
 
  • #4
And also once I messed up enough the hint was that
"A necessary condition for the potential to have a minimum is that its derivative is 0."

But the derivative of what? The only thing I can think of is the V(r)=kQ/r
 
  • #5
Potential at a point distance x from q1 is given by

V = k*q1/x + k*q2/(d-x) where d is the distance between the charges.

To find the minimum potential between the charge, find dV/dx and equate it to zero.
And find x.
 
  • #6
Thanks a lot! It was kind of a messy derivative but I am so thankful you helped me out. :)
 

FAQ: Find Potential Minimum of Two Point Charges

What is the formula for finding the potential minimum of two point charges?

The formula for finding the potential minimum of two point charges is V = k(Q1Q2)/r, where k is the Coulomb's constant, Q1 and Q2 are the magnitudes of the charges, and r is the distance between the charges.

How do the magnitudes and distances of the charges affect the potential minimum?

The potential minimum is directly proportional to the magnitudes of the charges and inversely proportional to the distance between them. This means that as the magnitudes of the charges increase, the potential minimum also increases, and as the distance between the charges increases, the potential minimum decreases.

Can the potential minimum be negative?

Yes, the potential minimum can be negative. This occurs when the charges have opposite signs and are close together. In this case, the potential energy is negative, indicating that the charges are attracted to each other.

How does the electric field affect the potential minimum?

The electric field is related to the potential minimum by the equation E = -∇V, where ∇V is the gradient of the potential. This means that the direction of the electric field is in the direction of decreasing potential, and the magnitude of the electric field is directly proportional to the rate of change of potential.

Can the potential minimum be used to determine the force between the charges?

Yes, the potential minimum can be used to determine the force between the charges. The force between two charges can be calculated using the equation F = k(Q1Q2)/r^2, where k is the Coulomb's constant, Q1 and Q2 are the magnitudes of the charges, and r is the distance between the charges. The potential minimum can then be used to calculate the potential energy, which is equal to the negative of the work done by the electric force.

Back
Top