Dielectric Slab Partially InsertedHardQuestion 31552
Question
A dielectric slab of thickness t (< d) with K is inserted in a parallel-plate capacitor (plate area A, separation d). Capacitance becomes:
- Aε₀A/(d−t+t/K)Correct
- Bε₀KA/(d−t)
- Cε₀A/(d+t(K−1))
- Dε₀KA/d
Correct answer
ε₀A/(d−t+t/K)
Explanation
The two air gaps (d−t) and dielectric (t) act as series capacitors. C = ε₀A/(d−t+t/K).