1. Problem Breakdown

We are asked to find the electric potential \( V_{\text{vertex}} \) at a vertex of a hollow cube made of 6 charged square plates.

Given:

• \( V_1 \): Potential at the center of a single square plate of charge \( q \).
• \( V_2 \): Potential at the center of the hollow cube formed by 6 such plates.

The vertex of the cube connects 3 adjacent plates, while the other 3 plates are “far” from this vertex.

2. Superposition Strategy

The potential at the vertex is the sum of contributions from the 3 “near” faces (which meet at the vertex) and the 3 “far” faces.

$$ V_{\text{vertex}} = 3V_{\text{near}} + 3V_{\text{far}} $$

Contribution of Near Faces (\( V_{\text{near}} \)):

Consider a square plate of side \( 2a \) made of 4 plates of side \( a \). The center of the large plate corresponds to the common corner (vertex) of the 4 constituent small plates.
Potential scales as \( V \propto \frac{Q}{L} \).
For the large plate (charge \( 4q \), side \( 2a \)): $$ V_{\text{center}, 2a} = k \frac{4q}{2a} \times (\text{geom factor}) = 2 \times V_{\text{center}, a} = 2V_1 $$ Due to symmetry, each of the 4 small plates contributes equally to the center potential. Thus, the potential at a corner of a square plate due to the plate itself is: $$ V_{\text{corner}} = \frac{2V_1}{4} = \frac{V_1}{2} $$ So, the 3 near faces contribute: $$ 3 V_{\text{near}} = 3 \left( \frac{V_1}{2} \right) = \frac{3V_1}{2} $$

3. Contribution of Far Faces using the “8-Cube” Method

To find the total potential, we can also use a scaling argument involving 8 such cubes stacked to form a larger cube of side \( 2L \). The center of this large cube corresponds to the vertex of the 8 constituent small cubes.

The potential at the center of the large structure (composed of 8 cubes) is \( 8 \times V_{\text{vertex}} \).

This large structure consists of:

1. An outer shell (a large hollow cube of side \( 2L \)). This is made of 24 plates.
2. An internal web of plates passing through the center. This consists of 3 large squares of side \( 2L \), each having double charge density (since plates from adjacent cubes touch).

Potential from Outer Shell:
The outer shell is a scaled-up version of the original cube (Side \( 2L \), Charge \( 24q \)). $$ V_{\text{outer}} = V_{\text{cube}} \times \frac{Q_{\text{total}}}{Q_{\text{orig}}} \times \frac{L_{\text{orig}}}{L_{\text{new}}} = V_2 \times \frac{24q}{6q} \times \frac{L}{2L} = V_2 \times 4 \times \frac{1}{2} = 2V_2 $$

Potential from Internal Web:
The internal web comprises 3 intersecting large squares of side \( 2L \). Each large square is made of 4 small plates, but has double charge (2 plates back-to-back). Total charge per large square = \( 4 \times 2q = 8q \).
Potential at center of a large square (Side \( 2L \), Charge \( 8q \)): $$ V_{\text{web\_plane}} = V_1 \times \frac{8q}{q} \times \frac{L}{2L} = V_1 \times 8 \times \frac{1}{2} = 4V_1 $$ There are 3 such planes, so total contribution is \( 3 \times 4V_1 = 12V_1 \).

Total Potential Equation:
Summing the contributions: $$ 8 V_{\text{vertex}} = V_{\text{outer}} + V_{\text{internal}} $$ $$ 8 V_{\text{vertex}} = 2V_2 + 12V_1 $$ Dividing by 8: $$ V_{\text{vertex}} = \frac{2V_2}{8} + \frac{12V_1}{8} $$ $$ V_{\text{vertex}} = \frac{V_2}{4} + \frac{3V_1}{2} $$