3D Dielectric Slab in Capacitor Simulation
Visualizing how a dielectric slab affects capacitance in 3D
Example 2.8
A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor but has a thickness (3/4)d, where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?
Initial Capacitor
Capacitance without dielectric:
C₀ = ε₀A/d
Electric field: E₀ = V₀/d
With Dielectric Slab
New capacitance:
C = (4K)/(K+3) × C₀
= 2.29 × C₀
Potential Difference Calculation
V = E₀(¼d) + (E₀/K)(¾d) = V₀(K+3)/4K
New potential: V = 0.44 × V₀
Key Concepts
- Dielectric reduces electric field by factor K
- Potential difference decreases when dielectric is inserted
- Capacitance increases due to dielectric
- Charge remains constant if capacitor is isolated
- Effect depends on both dielectric constant and thickness
