Solution:

To find the equivalent capacitance for this cubical capacitor network, where each edge of the cube has a capacitance

$C$

, here’s the shortest solution:

Step-by-Step:

1. Symmetry analysis:
   • By symmetry, all corners of the cube can be grouped into equivalent potential nodes.
   • The cube's symmetry allows reduction to a simpler circuit.
2. Key nodes:
   • Node
     $A$ 

     is connected to one corner of the cube.

   • Node
     $B$ 

     is connected to the diagonally opposite corner.

3. Effective connections:
   • Due to symmetry, three capacitors are effectively in parallel between
     $A$ 

     and an intermediate point.

   • Similarly, three capacitors are effectively in parallel between
     $B$ 

     and the same intermediate point.

   • Two capacitors remain directly between
     $A$ 

     and $B$ 

     .

4. Simplification:
   • The three parallel capacitors at each node result in:
     $$C_{\text{parallel}} = 3C$$ 
   • The equivalent circuit becomes two
     $3C$ 

     capacitors in series with a $2C$ 

     capacitor: $$\text{Series combination:} \quad \frac{1}{C_{\text{eq}}} = \frac{1}{3C} + \frac{1}{3C} + \frac{1}{2C}$$ 

5. Calculation:
   • Combine series:
     $$\frac{1}{C_{\text{eq}}} = \frac{2}{3C} + \frac{1}{2C} = \frac{4}{6C} + \frac{3}{6C} = \frac{7}{6C}$$ 
   • Invert to find
     $C_{\text{eq}}$ 

     : $$C_{\text{eq}} = \frac{6C}{7} \times 2 = \frac{12C}{7}$$ 

Thus, the equivalent capacitance is:

$$C_{\text{eq}} = \frac{12C}{7}$$