Potential at the corner of a insulated charged cube

[idon'tknowphy]

Homework Statement

An insulating cube of edge a has a uniform charge density p. The charge is zero everywhere outside the cube. The potential at an infinite distance from the cube is taken to be zero. If the potential at the center of the cube is Vo, find the potential at a corner of the cube. (Hint: Consider the potential at the center of a charged cube with the same charge density but with twice the length of the edge, use the principle of superposition in combination with a dimensional analysis.)

The problem is, I don't know how to find the potential at the center of a charged cube with length of the edge 2a

Homework Equations

i don't know if i am able to use gauss law to find the E-field out side and inside the cube

The Attempt at a Solution

i know that i can make a cube of side 2a by putting 4 cubes together, so the potential at the center of cube of side 2a will be
V_center of cube 2a = 8 V_corner

 

[Spinnor]

Make 7 copies of the charged cube of length a. We want to know the potential at the corner of the smaller cube, call that Vc. Arrange the 8 cubes into a larger cube of length 2a. We know that the potential at the center of the larger cube will be 8*Vc. We must now relate that to the known potential at the center of the smaller cube which is Vo somehow using dimensional analysis.

I'm stuck for now %^(

 

[idon'tknowphy]

i used Mathematica to calculate the potential at the center of the cube,
when a->2a, the potential will be 4 times larger

But i can't see it intuitively!

 

[Spinnor]

I'll throw this out and see if it makes sense.

Suppose we have cubes of uniform charge rho and of various size a. Let the potential at the center of such cubes be given by,

V = V(Q,R) where Q is the total charge on the cube and R is the length of the cubes edge.

Dimensionally

V = constant*Q/R

If this is true then if we double the edge length then Q --> 8Q and R --> 2*R so V --> 4V

Am I getting closer?

 

[asteriatic]

of cube a

I would approach this problem by first considering the basic principles of electrostatics. The potential at a point is defined as the work done per unit charge to bring a small positive test charge from infinity to that point. In this case, we can use the principle of superposition to find the potential at a corner of the cube by considering the potential due to each of the six faces of the cube individually.

We can start by considering a cube with edge length 2a, as suggested in the hint. The potential at the center of this cube can be calculated using the formula V = kq/r, where q is the total charge of the cube and r is the distance from the center. Since the charge density is uniform, we can use the formula for the total charge of a cube, q = pV, where V is the volume of the cube. In this case, V = (2a)^3 = 8a^3. Therefore, the potential at the center of the cube with edge length 2a is:

V_center of cube 2a = (k*p*8a^3)/(2a) = 4kpa^2

Now, we can use the principle of superposition to find the potential at the corner of the cube with edge length a. This means we need to consider the potential at the corner due to each of the six faces of the cube with edge length 2a. The distance from the corner to the center of each face is a, so we can use the formula V = kq/r, with q = p*a^3 for each face. This gives us a potential of:

V_corner of cube a = 6 (k*p*a^3)/a = 6kpa^2

Therefore, the potential at a corner of the cube is 6 times smaller than the potential at the center of a cube with twice the edge length. This makes sense, as the charge is spread out over a larger volume in the bigger cube, resulting in a smaller potential at a given distance.

In summary, the potential at a corner of the cube is V_corner of cube a = (1/6)V_center of cube 2a = (1/6)*4kpa^2 = (2/3)kpa^2. This solution was found using the principle of superposition and dimensional analysis.