Question

A slab of material of dielectric constant ‘K’ has the same area as the plates of a parallel plate capacitor but has a thickness $\frac{3}{4}d$, where d is the separation of the plates. The ratio of capacitance C (in the presence of dielectric) to the capacitance $C_0$ (in the absence of the dielectric) is :

• A. $\frac{3K}{K+4}$
  
• B. $\frac{3K}{K+1}$
• C. $\frac{4K}{K+3}$
• D. $\frac{4K}{3}$

Answer 1

The correct option is C $\frac{4K}{K+3}$


Electric field between the plates without dielectric, $E_0=\frac{\sigma }{{ϵ}_0}=\frac{q}{A{ϵ}_0}$

$\therefore$Potential difference between the plates,

$\Delta V=E_0\frac{d}{4}+\frac{E_0}{K}\times \frac{3d}{4}$

$\Delta V=\frac{E_{0}d}{4}(1+\frac{3}{K})$

$\Delta V=\frac{q.d}{4A{ϵ}_0}\left(\frac{K+3}{K}\right)$

New capacitance, $c^{\prime }=\frac{q}{\Delta V}=4c\left(\frac{K}{K+3}\right)$

$\frac{c^{\prime }}{c}=\frac{4K}{K+3}$