Step 1: Understanding the Concept:
This problem considers what happens when you introduce a conducting material into the electric field between the plates of a capacitor. Copper is an excellent metallic conductor. Inside any conductor in static equilibrium, the internal mobile electrons rearrange themselves perfectly to cancel out the external electric field, making the net electrostatic field inside the copper plate exactly zero ($E = 0$).
Step 2: Key Formula or Approach:
- The baseline capacitance of a vacuum-spaced parallel plate capacitor is:
\[ C_0 = \frac{\varepsilon_0 A}{d} \]
- When a conducting slab of thickness $b$ is added inside the plates, the original separation distance $d$ is effectively reduced because the field cannot exist within the thickness $b$. The electric field only spans across the remaining vacant space of thickness $(d - b)$.
- The potential difference ($V$) between the plates is given by $V = E_0 \times (d - b)$.
Step 3: Detailed Explanation:
Let's calculate the potential difference $V$ across the capacitor plates:
The electric field $E_0$ exists only in the air gaps. The total air space thickness is equal to the total distance minus the thickness of the copper metal:
\[ \text{Air Gap Thickness} = d - b \]
The potential difference $V$ is:
\[ V = E_0 (d - b) \]
We know that the electric field between two parallel plates with charge surface density $\sigma = \frac{Q}{A}$ is:
\[ E_0 = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A} \]
Substitute this field value back into our potential equation:
\[ V = \left(\frac{Q}{\varepsilon_0 A}\right) (d - b) \]
Using the definition of capacitance ($C = \frac{Q}{V}$):
\[ C = \frac{Q}{\left(\frac{Q}{\varepsilon_0 A}\right) (d - b)} = \frac{\varepsilon_0 A}{d - b} \]
This matches option (C).
Step 4: Final Answer:
The new capacity of the capacitor is $\frac{\varepsilon_0 A}{d - b}$.