1. Principle of Superposition and Scaling

We are dealing with a uniformly charged volume in the shape of a pyramid. We need to find the electric field \( E \) and potential \( V \) at the apex \( P \) after a top portion of height \( h \) has been removed.

The truncated pyramid can be modeled as:

Truncated Pyramid = Full Pyramid (Height H) – Small Pyramid (Height h)

Using the principle of superposition, the field and potential at point P are the differences between the values for the full pyramid and the values for the small removed pyramid.

2. Scaling Laws

For a uniform charge distribution of constant density \( \rho \), the electric field and potential scale with the linear dimension \( L \) as follows:

• Electric Field: \( E = \int k \frac{\rho d\tau}{r^2} \). Since \( d\tau \propto L^3 \) and \( r^2 \propto L^2 \), we have \( E \propto \frac{L^3}{L^2} \propto L \).
• Potential: \( V = \int k \frac{\rho d\tau}{r} \). Since \( d\tau \propto L^3 \) and \( r \propto L \), we have \( V \propto \frac{L^3}{L} \propto L^2 \).

3. Calculating Field and Potential

Let the field and potential of the full pyramid (height \( H \)) at the apex be \( E_0 \) and \( V_0 \).

For the small pyramid (height \( h \)):

Since it is geometrically similar to the large pyramid but scaled by a factor \( h/H \), its contributions at the apex are:

$$ E_{\text{small}} = E_0 \left( \frac{h}{H} \right) $$ $$ V_{\text{small}} = V_0 \left( \frac{h}{H} \right)^2 $$

For the truncated pyramid:

Subtract the contribution of the small pyramid from the full pyramid.

Electric Field:

$$ E = E_{\text{full}} – E_{\text{small}} = E_0 – E_0 \left( \frac{h}{H} \right) = E_0 \left( 1 – \frac{h}{H} \right) $$

Potential:

$$ V = V_{\text{full}} – V_{\text{small}} = V_0 – V_0 \left( \frac{h}{H} \right)^2 = V_0 \left( 1 – \frac{h^2}{H^2} \right) $$