Question

A parallel plate air capacitor has capacitance $C.$ Half of space between the plates is filled with di-electric di-electric constant $K$ as shown in figure. The new capacitance is $C^{\prime }$. Then

• A. $C^{\prime }=C\left[\frac{K}{K+1}\right]$
• B. $C^{\prime }=C\left[\frac{2K}{K+1}\right]$
• C. $C^{\prime }=\frac{2C}{K+1}$
• D. None of the above

Answer 1

Answer: (B)

$C^{\prime }=C\left[\frac{2K}{K+1}\right]$

The field with in the $1^{st}$ media

$E_1=\frac{\sigma }{{ϵ}_1}$

The field with in the $2^{nd}$ media

$E_2=\frac{\sigma }{{ϵ}_2}$

Potential Difference

$V=E_1{d}_1+E_2{d}_2$

$=\frac{\sigma }{{ϵ}_1}{d}_1+E_2{d}_2$

$=\sigma (\frac{d_1}{{ϵ}_1}+\frac{d_2}{{ϵ}_2})$

$=\frac{Q}{A}(\frac{d_1}{{ϵ}_1}+\frac{d_2}{{ϵ}_2})$

where $\sigma$ is the surface change density

$\sigma =\frac{Q}{A}$

$\therefore$ The capacitance in the case,

$C=\frac{Q}{V}=\frac{A}{\frac{d_1}{{ϵ}_1}+\frac{d_2}{{ϵ}_2}}=\frac{{ϵ}_{0}A}{\frac{d_1}{K_1}+\frac{d_2}{K_2}}$

where $K$ is are the dielectric constant

$\therefore {ϵ}_i=K_i{ϵ}_0$

If $K_1=1,K_2=K,e=\frac{{ϵ}_{0}A}{{\alpha }_1+\frac{d_2}{K}}[d_1=d_2=d]$

If $K_1=1,K_2=1,e=\frac{{ϵ}_{0}A}{2d}=\frac{K{ϵ}_{0}A}{d(K+1)}$

$\therefore e=e\left[\frac{2K}{K+1}\right]$