Question:

A parallel plate capacitor is constructed using two square plates of side length "s" and separation distance "d". The plates are made of a material with dielectric constant "k".

a) Derive an expression for the capacitance of this parallel plate capacitor.

b) The plates of the capacitor are connected to a battery with voltage V. When the plates are fully charged, the battery is disconnected and a dielectric slab with the same thickness as the separation distance is inserted between the plates. How does the capacitance change? Explain your answer.

c) The battery is then reconnected and the plates are recharged to their original voltage. How does the electric field between the plates change when the dielectric slab is inserted? Explain your answer.

Answer:

a) The capacitance of the parallel plate capacitor can be calculated using the following formula:

Where:

• C is the capacitance of the capacitor (in Farads, F)
• $\varepsilon_0$ is the vacuum permittivity ($8.85 \times 10^{-12} \, \text{F/m}$)
• A is the area of each plate (in $m^2$)
• k is the dielectric constant of the material between the plates (dimensionless)
• d is the separation distance between the plates (in meters)

b) When a dielectric slab is inserted between the plates, the capacitance of the capacitor increases. The capacitance can be calculated using the same formula as above, where $k$ is the dielectric constant of the material. Since the dielectric constant is greater than 1, it increases the overall capacitance of the capacitor.

Where $k_{\text{new}}$ is the dielectric constant of the material.

c) When the dielectric slab is inserted and the plates are recharged to their original voltage, the electric field between the plates decreases. The electric field can be calculated using the following formula:

Where:

• E is the electric field between the plates (in volts per meter, V/m)
• V is the voltage across the plates (in volts)
• d is the separation distance between the plates (in meters)

Since the distance remains the same while the dielectric constant increases, the electric field decreases. This is because the presence of the dielectric material reduces the electric field strength between the plates.