Question

Two thin dielectric slabs of dielectric constants $K_1$ and $K_2(K_1<K_2)$ are inserted between plates of a parallel plate capacitor, as shown in figure. The variation of electric field $E$ between the plates with distance $d$ as measured from plate $\text{P}$ is correctly shown by

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Let the electric field due to applied external source (battery) between the parallel plates of capacitor, be $E_0$.

After insertion of dielectric the electric field in the material of dielectric slab) present inside capacitor,

$E=\frac{E_0}{K}....\left(i\right)$

From, the given arrangement we infer that there are three regions in which dielectric material is not present.

Hence in these regions,
$E=E_0$

Given: $K_2>K_1$
Therefore in the two region which contains dielectric slabs,

$E_1=\frac{E_0}{K_1}=\text{constant}$

$E_2=\frac{E_0}{K_2}=\text{constant}$

$\therefore E_2<E_1<E_0$

Thus following graph is best representation of $E$ vs $d$.


Hence, option (c) is the correct answer.

$\begin{array}{c}\text{Why this question}?\\ \text{Tip: In such problems alway remember that "the}\\ \text{region with no dielectric will contain the same}\\ \text{electric which is produced by source However}\\ \text{the region with dielectric will undergo}\\ \text{reduction in}\overrightarrow{E}\text{value by k times"}\end{array}$