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. An 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 $\frac{Q}{K+}$
• D. Capacitance of the capacitor shown above is $(1+K)\frac{C_0}{2}$, where $C_0$ is the capacitance of the same capacitor with the dielectric removed

Answer 1

Answer: (C), (D)

Parallel

Capacitance of the capacitor shown above is

$C_{e}q=(1+k)\frac{C_0}{2}$

correct

Electric field in the dielectric-filled region is higher than that in the air-filled region - incorrect as $V\downarrow$

Charge on the half of the top plate above the air filled;

$Q_1=\frac{C_{0}Vk}{2}$

And, with air filled;

$Q_2=\frac{C_{0}V}{2}$ Potential is same

$\Rightarrow \frac{Q}{k}$

Charge on the half of the top plate above the air-filled part is - correct