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
$C_{1}=\frac{K \,\varepsilon_{0} \,A}{2 d}, C_{2}=\frac{\varepsilon_{0} \,A}{2 d}$
and $C_{ eq } =\frac{\varepsilon \cdot A}{2 d}(K+1) $
$ C_{ eq } =\frac{C_{0}}{2}(K+1) $
$ \frac{Q_{1}}{Q_{2}} =\frac{C_{1}}{C_{2}}=\frac{K}{1} $
$\Rightarrow \frac{\sigma_{1}}{\sigma_{2}} =\frac{K}{1}$
$Q_{1}=\frac{K Q}{K+1}$ and $Q_{2}=\frac{Q}{K+1}$
So, $E=\frac{\sigma}{\varepsilon_{0} K} $
$\Rightarrow \frac{E_{1}}{E_{2}}=\frac{\sigma_{1}}{\sigma_{2}} \times \frac{K_{1}}{1}$
$=\frac{Q_{1}}{Q_{2}} \times \frac{K_{2}}{K_{1}}=\frac{K}{1} \times \frac{1}{K}$
$=1: 1$