2020 Phase-2 Electrostatic Potential and CapacitanceCapacitor partially filled with dielectric slab Medium Question 33 of 76 in Electrostatic Potential and Capacitance

A parallel plate capacitor having cross-sectional area A and separation d has air in between the plates. Now an insulating slab of same area but thickness d/2 is inserted between the plates as shown in figure having dielectric constant K(= 4). The ratio of new capacitance to its original capacitance will be,

A 4 : 1
B 2 : 1
C 8 : 5
D 6 : 5

The answer is C.

Solution

Original capacitance $C_0=\dfrac{\varepsilon_0A}{d}$. With a slab of thickness $d/2$ and dielectric constant K inserted, new capacitance $C=\dfrac{\varepsilon_0A}{\dfrac{d}{2}+\dfrac{d/2}{K}}=\dfrac{\varepsilon_0A}{\dfrac{d}{2}\left(1+\dfrac{1}{K}\right)}$. With $K=4$: $C=\dfrac{\varepsilon_0A}{\dfrac{d}{2}\times\dfrac{5}{4}}=\dfrac{8\varepsilon_0A}{5d}$. Ratio $\dfrac{C}{C_0}=\dfrac{8}{5}$, i.e. 8:5.

Concept Capacitor partially filled with dielectric slab

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76 questions from this chapter have appeared across 27 exam years.