Work with Charges and Electric Potential

In summary, Homework Equations state that to assemble eight point charges, each of magnitude q, at the corners of a cube of side s, the equation becomes (keq/(1/2)(sqrt3)s). The attempt at a solution calculated that 8(keq/(1/2)(sqrt3)s) ... where 8 is to incorporate the fact there are 8 equal charges all the same s away.
  • #1
Alex G
24
0

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?
 
Physics news on Phys.org
  • #2
Notice there are 28 distances from corner to corner on a cube.
 
  • #3
I am not quite seeing 28 distances, in fact I can only really imagine the 8 points here.
 
  • #4
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.
 
  • #5
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).
 

1. What is the difference between electric potential and electric potential energy?

Electric potential is the measure of the electric potential energy per unit charge at a given point in space. In other words, it is the electric potential energy that a unit charge would have if placed at that point. Electric potential energy, on the other hand, is the energy a system of charges has due to their positions and interactions with each other.

2. How are electric potential and electric field related?

The electric potential at a point in space is directly related to the electric field at that point. The electric field is a vector quantity that describes the force experienced by a unit charge at a given point. Electric potential, on the other hand, is a scalar quantity that describes the potential energy per unit charge at a given point. The electric field is the negative gradient of the electric potential, meaning that the direction of the electric field is opposite to the direction of increase in electric potential.

3. Can the electric potential at a point be negative?

Yes, the electric potential at a point can be negative. This means that the potential energy per unit charge at that point is negative, indicating that the charge would have lower potential energy at that point compared to a reference point. However, the magnitude of the electric potential is what determines the strength of the electric field, so a negative potential does not necessarily mean a weaker electric field.

4. How is electric potential calculated for a system of charges?

The electric potential at a point due to a system of charges is the sum of the electric potentials due to each individual charge. This can be calculated using the equation V = kQ/r, where V is the electric potential, k is the Coulomb's constant, Q is the charge of the object creating the potential, and r is the distance from the object to the point where the potential is being measured. If there are multiple charges, the electric potentials due to each charge are added together.

5. What is the significance of equipotential lines in electric potential?

Equipotential lines are imaginary lines that connect points with the same electric potential. They are perpendicular to electric field lines and indicate that there is no work being done in moving a charge along that line. These lines are important in visualizing the electric field and understanding the behavior of charges in an electric field. The closer the equipotential lines are to each other, the stronger the electric field at that point.

Similar threads

Replies
22
Views
1K
  • Introductory Physics Homework Help
Replies
23
Views
345
Replies
1
Views
144
  • Introductory Physics Homework Help
Replies
6
Views
161
  • Introductory Physics Homework Help
Replies
1
Views
890
  • Introductory Physics Homework Help
Replies
2
Views
364
  • Introductory Physics Homework Help
Replies
1
Views
151
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
201
  • Introductory Physics Homework Help
Replies
17
Views
400
Back
Top