Solution
We use the Principle of Superposition. The new structure is equivalent to the original full cube (side $a$) minus a smaller cube (side $a/2$) located at the corner P.
1. Scaling Laws:
For a uniformly charged object of characteristic length $L$ and density $\rho$:
- Potential $V \propto L^2$
- Electric Field $E \propto L$
2. Calculations:
- Original Cube (Side $a$):
- $V_{total} = V_0$
- $E_{total} = E_0$ (Directed along the body diagonal outward from center)
- Removed Cube (Side $a/2$):
Point P is also a corner of this smaller cube.
- $V_{removed} = V_0 \times (1/2)^2 = V_0 / 4$
- $E_{removed} = E_0 \times (1/2) = E_0 / 2$ (Also directed along the body diagonal)
3. Resultant Values:
- Potential (Scalar): $$V_{net} = V_{total} – V_{removed} = V_0 – \frac{V_0}{4} = \frac{3V_0}{4}$$
- Electric Field (Vector):
Since both the original cube and the removed corner cube share the same diagonal axis passing through P, their field vectors are collinear. We can subtract the magnitudes directly. $$E_{net} = E_{total} – E_{removed} = E_0 – \frac{E_0}{2} = \frac{E_0}{2}$$
Answer: (a) $E_0/2$ and $3V_0/4$