Calculate the capacitance and free charge on the plates of a parallel-plate capacitor with a dielectric slab.

Here's a step-by-step solution to the problem:

First, let's list the given values and convert them to SI units:

• Plate area, $A = 90 \text{ cm}^2 = 90 \times (10^{-2} \text{ m})^2 = 9 \times 10^{-3} \text{ m}^2$
• Plate separation, $d = 1.14 \text{ cm} = 1.14 \times 10^{-2} \text{ m}$
• Potential difference, $V = 60 \text{ V}$
• Dielectric slab thickness, $t = 0.6 \text{ cm} = 0.6 \times 10^{-2} \text{ m} = 6 \times 10^{-3} \text{ m}$
• Dielectric constant, $K = 2.7$
• Permittivity of free space, $\epsilon_0 = 8.85 \times 10^{-12} \text{ F/m}$

a) The capacitance of the capacitor.

Step 1: Use the formula for the capacitance of a parallel plate capacitor in a vacuum (or air, which is a good approximation).

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

Step 2: Substitute the given values into the formula.

$$C = \frac{(8.85 \times 10^{-12} \text{ F/m}) \times (9 \times 10^{-3} \text{ m}^2)}{1.14 \times 10^{-2} \text{ m}}$$

Step 3: Calculate the capacitance.

$$C = \frac{7.965 \times 10^{-14} \text{ F}\cdot\text{m}}{1.14 \times 10^{-2} \text{ m}}$$ $$C = 6.9868 \times 10^{-12} \text{ F}$$

The capacitance is approximately $6.99 \text{ pF}$.

$$\boxed{\text{C} = 6.99 \times 10^{-12} \text{ F}}$$

b) What free charge appears on the plate.

Step 1: Use the relationship between charge, capacitance, and potential difference.

$$Q = CV$$

Step 2: Substitute the calculated capacitance from part (a) and the given potential difference.

$$Q = (6.9868 \times 10^{-12} \text{ F}) \times (60 \text{ V})$$

Step 3: Calculate the charge.

$$Q = 4.19208 \times 10^{-10} \text{ C}$$

The free charge on the plate is approximately $4.19 \times 10^{-10} \text{ C}$.

$$\boxed{\text{Q} = 4.19 \times 10^{-10} \text{ C}}$$

c) If the battery was disconnected and a dielectric slab of thickness $0.6 \text{ cm}$ was inserted between the plates. What is the electric field in the gap between the plates and the dielectric slab?

When the battery is disconnected, the charge $Q$ on the plates remains constant. The electric field will be different in the air gap and inside the dielectric slab.

Step 1: Calculate the electric field in the air gap (the region without the dielectric). The electric field in the air gap is determined by the surface charge density $\sigma = Q/A$.

$$E_{air} = \frac{\sigma}{\epsilon_0} = \frac{Q}{A\epsilon_0}$$

Step 2: Substitute the charge from part (b) and the given values.

$$E_{air} = \frac{4.19208 \times 10^{-10} \text{ C}}{(9 \times 10^{-3} \text{ m}^2) \times (8.85 \times 10^{-12} \text{ F/m})}$$ $$$$