A parallel plate capacitor has a plate separa | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The capacitance $C$ of a parallel plate capacitor with a dielectric slab of thickness $t$ and dielectric constant $K$ is given by the formula:$C =

Vedclass · Accepted Answer

$\frac{\varepsilon_0 A}{d - t(1 - 1/K)}$

Vedclass · Answer

(A) The capacitance $C$ of a parallel plate capacitor with a dielectric slab of thickness $t$ and dielectric constant $K$ is given by the formula:$C = \frac{\varepsilon_0 A}{d - t + t/K}$This can be rearranged as:$C = \frac{\varepsilon_0 A}{d - t(1 - 1/K)}$Here,the effective distance between the plates is reduced due to the presence of the dielectric slab. The potential difference $V$ across the plates is the sum of the potential drop across the air gap $(d-t)$ and the dielectric slab $t$:$V = E_{	ext{air}}(d-t) + E_{	ext{medium}}(t)$Since $E_{	ext{air}} = \frac{\sigma}{\varepsilon_0}$ and $E_{	ext{medium}} = \frac{\sigma}{K\varepsilon_0}$,where $\sigma = \frac{Q}{A}$:$V = \frac{Q}{A\varepsilon_0}(d-t) + \frac{Q}{A\varepsilon_0 K}(t) = \frac{Q}{A\varepsilon_0} [d - t + t/K] = \frac{Q}{A\varepsilon_0} [d - t(1 - 1/K)]$Using $C = \frac{Q}{V}$,we get $C = \frac{\varepsilon_0 A}{d - t(1 - 1/K)}$.

$A$ parallel plate capacitor has a plate separation of $d$ and each plate has an area $A$. What is the capacitance when a dielectric slab of thickness $t$ and dielectric constant $K$ is placed between the plates?

Similar Questions

Three parallel plate capacitors each with area $A$ and separation $d$ are filled with two dielectrics ($k_1$ and $k_2$) in the following fashion. Which of the following is true? $(k_1 > k_2)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module