1) Both the Coulomb law and gravitational forces lead to potential energies of the form [see the attached file for full equations and symbols] where [...] denotes [...] in the case of the Coulomb force and [...] for gravity, and r1 and r2 are the positions of the two particles. Show in detail that [...] is the force on particle 1 and [...] that on particle 2.
2) Write out the arguments of all the potential energies of the four-particle system in
For instance U = U (r1, r2, ... , r4), whereas U_34 = U_34 (r3 ??" r4). Show in detail that the net force on particle 3 (for instance) is given by [...].
Please see the attachment for the full questions.© BrainMass Inc. brainmass.com September 19, 2018, 3:34 pm ad1c9bdddf - https://brainmass.com/physics/gamma/gradient-of-potential-energy-equals-minus-the-force-106846
You can do problem 1) in Cartesian coordinates or in spherical coordinates. Let me give an outline of how you can do it in Cartesian coordinates first.
The potential energy is:
U = gamma/|r1 - r2|
We will calculate minus the gradient and then compare that with the force law to see if minus the gradient is indeed the force. In Cartesian coordinates you can write:
|r1 - r2| = sqrt[(x1-x2)^2 + (y1 - y2)^2 + (z1 - z2)^2]
The components of the gradient are the derivatives with respect to the coordinates. You can take the gradient w.r.t. to r1 = (x1,y1,z1) or r2 = (x2,y2,z2). If we calculate the gradient w.r.t. r1 and take the x-component then we have to differentiate U w.r.t. x1. Let's first rewrite U as:
U = gamma [(x1-x2)^2 + (y1 - y2)^2 + (z1 - z2)^2]^(-1/2)
dU/dx1 = -1/2gamma[(x1-x2)^2 + (y1 - y2)^2 + (z1 - z2)^2]^(-3/2)*2(x1-x2) =
-gamma[(x1-x2)^2 + (y1 - y2)^2 + (z1 - z2)^2]^(-3/2)*(x1-x2).
The term in the square brackets is |r1-r2|^2, so yo can write this in a "mixed" form as:
dU/x1 = -gamma/|r1-r2|^3 (x1-x2) ...