Question

An isolated large-plate capacitor (not connected to anything) originally has a potential difference of 930 volts with an air gap of 2 mm. Then a plastic slab 1 mm thick, with dielectric constant 5.5, is inserted into the middle of the air gap as shown in the figure. As shown in the diagram, location 1 is at the left plate of the capacitor, location 2 is at the left edge of the plastic slab, location 3 is at the right edge of the slab, and location 4 is at the right plate of the capacitor. All of these locations are near the center of the capacitor. Calculate the following potential differences.
V1 - V2 =   V
V2 - V3 =  V
V3 - V4 =  V
V1 - V4 =  V

Answer 1

According to the given problem,

Assuming radius of the plates is much larger than the distance between the plates, end effects are negligible and there is a uniform electrical field for each section of the capacitor.
E = ΔV / Δs

The electrical field in the capacitor without plastic slab is:
E₀ = 930V / 2×10⁻³m = 4.65×10⁵Vm⁻¹

After inserting of the slab the electrical field inside the air gaps is the same. Within the slab it decreases to
E = E₀/ε = 4.65×10⁵Vm⁻¹/5.5 = 0.85×10⁵Vm⁻¹

Multiply electrical field by the distance to calculate the potential difference:
ΔV = E·Δs

Hence:
V₁-V₂ = 4.65×10⁵Vm⁻¹ · 0.5×10⁻³m = 232.5V
V₂-V₃ = 0.85×10⁵Vm⁻¹ · 1×10⁻³m = 85V
V₃-V₄ = 4.65×10⁵Vm⁻¹ · 0.5×10⁻³m = 232.5V
=>
V₁-V₄ = (V₁-V₂) + (V₂-V₃) + (V₃-V₄)
= 232.5V + 85V + 232.5V = 550V

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