If the aforementioned capacitor is connected across a 6.0 V...

The induced charge on the dielectric is given by the formula:

q_{ind} = Q \left(1 -\dfrac{1}{K}\right)

where $Q$ is the total charge on the capacitor plates and $K$ is the dielectric constant of the material.

When the capacitor is connected across a battery of voltage $V$, the charge on the plates with the dielectric inserted is:

$Q = K C_0 V$

Substituting this into the induced charge formula, we get:

q_{ind} = K C_0 V \left(1 -\dfrac{1}{K}\right) =$C_0 V (K - 1)$

Based on standard problem parameters (e.g., a parallel plate capacitor with area $A = 400 \text{ cm}^2$, separation $d = 1.0 \text{ mm}$, and dielectric constant $K = 5$), the initial capacitance $C_0$ is:

$C_0 =$\dfrac{\epsilon_0 A}{d} = \dfrac{8.854 \times 10^{-12} \times 400 \times 10^{-4}}{1.0 \times 10^{-3}} =$354.16 \text{ pF}$

Using the given battery voltage $V = 6.0 \text{ V}$:

$q_{ind} = 354.16 \times 10^{-12} \times 6.0 \times (5 - 1)$

$q_{ind} = 354.16 \times 10^{-12} \times 24 = 8499.8 \times 10^{-12} \text{ C}$

$q_{ind} \approx 8.5 \text{ nC}$

Answer: $8.5 \text{ nC}$