2. Electric Potential and Capacitance
medium

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 the figure. The variation of electric field $E$ between the plates with distance $d$ as measured from plate $P$ is correctly shown by

A
B
C
D
(AIPMT-2014)

Solution

Electric field inside parallel plate capacitor having charge $Q$ at place where dielectric is absent $=\frac{Q}{A\varepsilon_0}$

where dielectric is present $=\frac{Q}{A\varepsilon_0}$ 

As $K_1 < K_2\, so\, E_1 > E_2$

Hence graph $(c)$ correctly depicts the variation of electric field $E$ with distance $d.$

Standard 12
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.