Work with Charges and Electric Potential

AI Thread Summary
The discussion revolves around calculating the work required to assemble eight identical point charges at the corners of a cube. The initial approach involved determining the distance from the center of the cube to the corners and using the electric potential formula, but the calculations were found to be incorrect. Participants suggest a more systematic method of bringing in charges one at a time, calculating the potential energy for each addition, and summing the total work done. An alternative method involves calculating the force on one charge and integrating the work as the cube is shrunk from infinity to its final size. The complexity of the problem lies in accurately accounting for the interactions between multiple charges and ensuring the final answer reflects potential energy.
Alex G
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Homework Statement



How much work is required to assemble eight identical point charges, each of magnitude q, at the corners of a cube of side s? Note: Assume a reference level of potential V = 0 at r = infinity. (Use any variable or symbol stated above along with the following as necessary: ke.)

Homework Equations



Some that may be useful are
-Work = Delta U = q Delta V

Where U = potential energy and V= Electric Potential

The Attempt at a Solution



I started by first putting the "center" of the cube at the origin of a 3D plane. From there after some Pythagorean Theorem I determined that the origin is (1/2)(sqrt3)s away from all charges q at each corner.
Noting that the initial potential difference is 0 the equation became (for these point charges) (keq/r) Where r = s in this instance.
Thus my answer became 8(keq/(1/2)(sqrt3)s) ... where 8 is to incorporate the fact there is 8 equal charges all the same s away.
I feel this is correct, but it is saying it is wrong. Any confirmation or corrections?
 
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Notice there are 28 distances from corner to corner on a cube.
 
I am not quite seeing 28 distances, in fact I can only really imagine the 8 points here.
 
The usual way these problems are approached is to "bring in" one charge at a time (the first one's "free") and calculate the work required to do so. So the second charge brought in "sees" the first charge, the third charge "sees" the first two, and so on. The number of potentials that need to be calculated goes up by one for each new charge brought in. Sum up the total and you're done.

An alternative would be to calculate the force on one charge of a cube of side s and note that, by symmetry, the force is directed outward along the radial line from the cube center to that charge. All the other charges at their vertexes will experience the same magnitude of force, but by symmetry, directed outward in the appropriate radial directions.

Express the force in terms of the radial distance r (replace "s" with an expression in r). Now suppose that you let r become very large. The cube will similarly grow large. In this limit r goes to infinity and the charges are all at infinity. Calculate the work required to shrink the cube back down to side size a by integrating 8x the force as r goes from infinity down to the appropriate length (half the cube diagonal length). Still a lot of work (math) but perhaps less tedious than doing the charges one at a time.
 
Thanks gneil. I agree on both of those approaches. I was attempting that first one to try and avoid any integrals. I kept getting lost after determining which charge has experienced what. Eventually I was able to determine with that same idea that each point charge is receiving in terms of kq(1/r + 1/r2 + 1/r3 + ...) (because of electric potential equation when q is constant): 3(1/s + 1/sqrt2*s) + (1/sqrt3*s) all * kq ... because 3 charges act each in those first 2 manners and one charge in the 3rd manner. All multiplied by 8 to account for all 8 charges. And since Delta V = Delta U/q ... the answer was then multiplied to make it q2 (they wanted the answer in determines of potential energy, not W specifically).
 
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