Question

Half of the space between the plates of a parallel-plate capacitor is filled with a dielectric material of dielectric constant $K$. The remaining half contains air as shown in the figure. The capacitor is now given a charge $Q$. Then

• A. electric field in the dielectric-filled region is higher than that in the air-filled region
• B. on the two halves of the bottom plate the charge densities are unequal
• C. charge on the half of the top plate above the air-filled part is $\frac{Q}{K+1}$
• D. capacitance of the capacitor shown above is $\left(1+K\right)\frac{C_0}{2}$, where $C_0$ is the capacitance of the same capacitor with the dielectric removed

Answer 1

Answer: (B), (C), (D)

We know that
$C_1=\frac{K{\epsilon }_{0}A}{2d},C_2=\frac{{\epsilon }_{0}A}{2d}$
and $C_{eq}=\frac{\epsilon \cdot A}{2d}(K+1)$
$C_{eq}=\frac{C_0}{2}(K+1)$
$\frac{Q_1}{Q_2}=\frac{C_1}{C_2}=\frac{K}{1}$
$\Rightarrow \frac{{\sigma }_1}{{\sigma }_2}=\frac{K}{1}$
$Q_1=\frac{KQ}{K+1}$ and $Q_2=\frac{Q}{K+1}$
So, $E=\frac{\sigma }{{\epsilon }_{0}K}$
$\Rightarrow \frac{E_1}{E_2}=\frac{{\sigma }_1}{{\sigma }_2}\times \frac{K_1}{1}$
$=\frac{Q_1}{Q_2}\times \frac{K_2}{K_1}=\frac{K}{1}\times \frac{1}{K}$
$=1:1$