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2. Electric Potential and Capacitance
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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