Question

A parallel plate capacitor has a dielectric slab of dielectric constant $K$ between its plate that covers $1/3$ of the area of its plates, as shown in the figure. The total capacitance of the capacitors is $C$, while that of the portion with the dielectric in between is $,C_1$. When the capacitor is charged, the plate area covered by the dielectric gets to charge $Q_1$ and the rest of the area gets to charge $Q_2$. The electric field in the dielectric is $E_1$ and that in the other portion is $E_2$. Choose the correct option/options, ignoring the edge effects.

• A. $\frac{E_1}{E_2}=1$
• B. $\frac{E_1}{E_2}=\frac{1}{K}$
• C. $\frac{Q_1}{Q_2}=\frac{3}{K}$
• D. $\frac{C}{C_1}=\frac{2+K}{K}$

Answer 1

Answer: (A), (D)

The correct options are
A $\frac{E_1}{E_2}=1$
D $\frac{C}{C_1}=\frac{2+K}{K}$

$\Rightarrow C_1=\frac{\left(\frac{A}{3}\right)E_{0}k}{d}=\frac{A{E}_{0}k}{3d}$

$\Rightarrow C=C_1+C_2$

$=\frac{A{E}_{0}k}{3d}(k+2)$

$\Rightarrow C_2=\frac{\left(\frac{2A}{3}\right)E_0}{d}=\frac{2A{E}_0}{3d}$

$\Rightarrow$ Option (D) $\frac{C}{C_1}=\frac{\frac{A{E}_{0}k}{3d}(k+2)}{\frac{A{E}_{0}k}{3d}}=\frac{k+2}{k}$

Now, for both the regions, potential difference across the plates of the capacitors is same. ( let $=v$ )

$\Rightarrow E_1.d=E_2.d=v$

$\Rightarrow E_1=E_2$

$\Rightarrow \frac{E_1}{E_2}=1$ (A)

Hence, both option (A) and (D) are correct.