Solution:

The question involves calculating the capacitance of a parallel-plate capacitor when a dielectric slab of thickness

$t$

and dielectric constant

$K = 2$

is partially inserted between the plates.

The short solution is based on treating the system as a combination of two capacitors in series:

1. Capacitor 1 (region with dielectric): The thickness of this region is
   $t$ 

   , and the capacitance is given by: 

   $$C_1 = \frac{\varepsilon_0 A K}{t} = \frac{2\varepsilon_0 A}{t}$$ 

2. Capacitor 2 (region without dielectric): The thickness of this region is
   $d - t$ 

   , and the capacitance is: 

   $$C_2 = \frac{\varepsilon_0 A}{d - t}$$ 

Since the two regions are in series, the equivalent capacitance is given by:

$$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$$

Substituting

$C_1$

and

$C_2$

:

$$\frac{1}{C} = \frac{t}{2\varepsilon_0 A} + \frac{d - t}{\varepsilon_0 A}$$

Simplifying:

$$\frac{1}{C} = \frac{t + 2(d - t)}{2\varepsilon_0 A} = \frac{2d - t}{2\varepsilon_0 A}$$

Therefore:

$$C = \frac{2\varepsilon_0 A}{2d - t}$$

When the dielectric slab thickness is

$t = \frac{d}{2}$

, the capacitance simplifies to:

$$C = \frac{\varepsilon_0 A}{d - \frac{t}{2}}$$

This matches the given answer. Let me know if further clarification is needed!