Step 1: Calculate the initial capacitance of the parallel plate capacitor.

When the medium between the plates is air (or vacuum), the capacitance ($C_0$) is given by:

$$C_0 = \frac{\epsilon_0 A}{d}$$

where:

$\epsilon_0$ is the permittivity of free space.
$A$ is the area of each plate.
$d$ is the separation between the plates.

Step 2: Calculate the capacitance after introducing the dielectric slab.

When a dielectric slab of thickness $t$ and dielectric constant $K$ is introduced between the plates of a parallel plate capacitor, the new capacitance ($C_f$) is given by the formula:

$$C_f = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$$

In this problem, the dielectric constant $K = 3$. So, substituting this value:

$$C_f = \frac{\epsilon_0 A}{d - t + \frac{t}{3}}$$

To simplify the denominator:

$$C_f = \frac{\epsilon_0 A}{d - \frac{3t}{3} + \frac{t}{3}}$$

$$C_f = \frac{\epsilon_0 A}{d - \frac{2t}{3}}$$

Step 3: Use the given condition about the increase in capacity.

The problem states that the capacity increases by $50 \%$. This means the final capacitance $C_f$ is $150\%$ of the initial capacitance $C_0$.

$$C_f = C_0 + 50\% \text{ of } C_0$$

$$C_f = C_0 + 0.5 C_0$$

$$C_f = 1.5 C_0$$

Step 4: Equate the expressions for $C_f$ and solve for the thickness $t$.

Now, substitute the expressions for $C_f$ and $C_0$ into the equation from Step 3:

$$\frac{\epsilon_0 A}{d - \frac{2t}{3}} = 1.5 \left( \frac{\epsilon_0 A}{d} \right)$$

We can cancel out $\epsilon_0 A$ from both sides of the equation:

$$\frac{1}{d - \frac{2t}{3}} = \frac{1.5}{d}$$

Now, cross-multiply to solve for $t$:

$$d = 1.5 \left( d - \frac{2t}{3} \right)$$

$$d = 1.5d - 1.5 \times \frac{2t}{3}$$

$$d = 1.5d - \frac{3}{2} \times \frac{2t}{3}$$

$$d = 1.5d - t$$

Rearrange the equation to isolate $t$:

$$t = 1.5d - d$$

$$t = 0.5d$$

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

Thus, the thickness of the dielectric slab is $\frac{d}{2}$.

The final answer is $\boxed{\frac{\mathrm{d}}{2}}$.

Capacitor

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