In a parallel plate air capacitor of plate separation $d$, a dielectric slab of thickness $t$ is introduced between the plates $(t

Question

In a parallel plate air capacitor of plate separation $d$, a dielectric slab of thickness $t$ is introduced between the plates $(t < d)$. The capacitance becomes one-third of the original value. The dielectric constant of the slab will be

Aicharya · Accepted Answer

Equivalent Capacitance of the dielectric-air system is given by $\frac{1}{C^{\prime}}=\left(\frac{d-t}{\varepsilon_0 A}ight)+\left(\frac{t}{K \varepsilon_0 A}ight)=\frac{d-t\left(1-\frac{1}{K}ight)}{\varepsilon_0 A}$ If new capacitance is given by one third of the old value: $\begin{aligned} & C=\frac{\varepsilon_0 A}{d-t\left(1-\frac{1}{K}ight)}=\frac{\varepsilon_0 A}{3 d} \ & \Rightarrow d-t+\frac{t}{K}=3 d \ & \Rightarrow K=\frac{t}{2 d+t} \end{aligned}$

In a parallel plate air capacitor of plate separation $d$, a dielectric slab of thickness $t$ is introduced…

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