The image you uploaded contains an example related to the capacitance of a parallel plate capacitor with a dielectric medium between the plates. Here's a breakdown of the problem:

Given:

• Dielectric thickness $d = 0.5 \, \text{cm}$
• Relative permittivity $\epsilon_r = 3$
• Plate dimensions: $\text{length} = 10 \, \text{cm}, \text{width} = 15 \, \text{cm}$
• Electric field strength in the dielectric medium $E = 30 \, \text{V/cm}$

We are tasked with determining:

1. The capacitance of the capacitor.
2. The total charge on each plate.

Formulae:

1. Capacitance of a parallel plate capacitor: $C = \epsilon_0 \epsilon_r \frac{A}{d}$ Where:

   • $\epsilon_0$ is the vacuum permittivity, approximately $8.85 \times 10^{-12} \, \text{F/m}$.
   • $\epsilon_r$ is the relative permittivity of the dielectric.
   • $A$ is the area of the plates.
   • $d$ is the separation between the plates.

2. Total charge on the plates: $Q = C \cdot V$ Where $V$ is the potential difference across the plates.

I will proceed with the calculations now. Constants epsilon_0 = 8.85e-12 # Vacuum permittivity in F/m epsilon_r = 3 # Relative permittivity d = 0.5e-2 # Dielectric thickness in meters (converted from cm) length = 10e-2 # Length of the plate in meters (converted from cm) width = 15e-2 # Width of the plate in meters (converted from cm)

Area of the plates

A = length * width

Capacitance formula

C = epsilon_0 * epsilon_r * A / d

Given electric field strength

E = 30 # in V/cm V = E * d * 100 # Convert electric field to volts and calculate potential difference

Total charge

Q = C * V

C, QThe calculations for the given capacitor are as follows:

1. Capacitance of the parallel plate capacitor is approximately: $C \approx 7.97 \times 10^{-11} \, \text{F} \, (\text{or} \, 79.7 \, \text{pF})$

2. The total charge on each plate is approximately: $Q \approx 1.19 \times 10^{-9} \, \text{C} \, (\text{or} \, 1.19 \, \text{nC})$

Would you like further details or clarifications? Here are some related questions that may help expand on this:

1. How does the dielectric constant affect the capacitance of a capacitor?
2. What happens to the capacitance if the dielectric thickness is doubled?
3. How would you calculate the potential difference between plates using a different dielectric?
4. What is the effect of using a vacuum as the dielectric material in this setup?
5. How does the charge on the capacitor change if the area of the plates is increased?

Tip: When dealing with parallel plate capacitors, always ensure that the units of length, width, and dielectric thickness are consistent for accurate results.