Potential anywhere inside a cube

In summary: The solution to the full problem is then the sum of the solutions to the individual parts.In summary, the problem involves finding the electrostatic potential ##\Phi(x,y,z)## at any point inside a cube with six different potentials on each face. The solution involves breaking the problem into several parts and solving each part separately, then combining the solutions to find the full solution. This involves finding the constants ##A, B, C, D, E, F## and the values of ##k_i##.
  • #1
andre220
75
1

Homework Statement


All six faces of a cube, of side length ##L##, are maintained at constant, but different potentials. The left and right faces are at ##V_1## each. The top and bottom are at ##V_2## each. The front and back are at ##V_3##. Determine the electrostatic potential ##\Phi(x,y,z)## at any point inside the cube. What is the value of the electrostatic potential at the center of the cube?

Homework Equations


Solutions are of the form:

##X(x) = A \sin{k_1 x} + B\cos{k_1 x}##
##Y(y) = C\sin{k_2 y} + D\cos{k_2 y}##
##Z(z) = E\sinh{k_3 z} + F\cosh{k_3 z}##
where ##k_3 = \sqrt{k_1 + k_2}##

The Attempt at a Solution


Normally, the case in which one or two sides are held at some potential, the symmetry is exploited to get ##k_i## and the prevailing constants. However, here, that does not seem to be possible here.

When: ##x =0, x= L##, ##\Phi = V_1##
##y =0, y= L##, ##\Phi = V_2##
##z =0, z= L##, ##\Phi = V_3##

However, I do not see how this get me any closer to finding, any of the constants ##A,B,C,D,E,F##. ##\Phi = X(x)Y(y)Z(z)##.
Thus for the first B.C.:

##\Phi(0,y,z) = (A\sin{0} + B\cos{0})(C\sin{k_2 y} + D \cos{k_2 y})(E\sinh{k_3 z} + F\cosh{k_3 z})##
##= B (C\sin{k_2 y} + D \cos{k_2 y})(E\sinh{k_3 z} + F\cosh{k_3 z}) = V_1##

and so do I just keep doing this and will eventually, get my coefficients? Regardless, I don't necessarily see how ##k_i## would be found.

Thank you
 
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  • #2
The problem is a linear PDE. You can solve for one inhomogeneity at a time and make a superposition of the solutions to find the full solution.
 
  • #3
Okay, I take when ##x=0\implies## ##V_1##:
##A\sin{k_x 0} + B\cos{k_x 0} = V_1 \implies B = V_1## and I could keep going on like that for each of the six boundary conditions
But I am still not seeing how that would work. Plus I don't see how I could get the ##k##'s from this.
 
  • #4
No. You need to split the entire problem into several parts, then solve each part separately, including the now homogeneous boundary conditions on the sides.
 

Related to Potential anywhere inside a cube

1. What is the potential inside a cube?

The potential inside a cube is the measure of the electrostatic potential energy of a point charge placed at any location inside the cube. It is a scalar quantity that depends on the position and charge of the point charge.

2. How is the potential inside a cube calculated?

The potential inside a cube can be calculated using the formula V = kq/r, where V is the potential, k is the Coulomb's constant, q is the point charge, and r is the distance between the point charge and the location inside the cube.

3. Is the potential inside a cube constant?

No, the potential inside a cube is not constant. It varies depending on the location inside the cube and the charge of the point charge.

4. Can the potential inside a cube be negative?

Yes, the potential inside a cube can be negative. This indicates that the point charge is surrounded by a region of negative potential energy, which means that the force acting on the point charge is attractive.

5. How does the potential inside a cube change with distance?

The potential inside a cube follows an inverse relationship with distance. As the distance between the point charge and the location inside the cube increases, the potential decreases. This is because the electrostatic force between two charges decreases with distance.

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