A parallel-plate capacitor with a capacitance of 5 is connected to...

The capacitance of the original parallel-plate capacitor is given by:

$C_0 = \dfrac{\varepsilon_0 A}{d} = 5 \mu\text{F}$

When a dielectric slab of thickness $t$ and dielectric constant $K$ is introduced between the plates, the new capacitance is:

$C = \dfrac{\varepsilon_0 A}{d - t + \dfrac{t}{K}}$

Given $d = 2 \text{ mm}$, $t = 1 \text{ mm}$, and $K = 5$, we substitute these values into the formula:

$C = \dfrac{\varepsilon_0 A}{2 - 1 + \dfrac{1}{5}} = \dfrac{\varepsilon_0 A}{1.2}$

From the initial condition, we have $\varepsilon_0 A = C_0 \times d = 5 \times 2 = 10 \mu\text{F}\cdot\text{mm}$.

Substituting this into the expression for the new capacitance:

$C = \dfrac{10}{1.2} = \dfrac{100}{12} = \dfrac{25}{3} \approx 8.3 \mu\text{F}$