Q.Two thin dielectric slabs of dielectric constant $K_1$ and $K_2$ are inserted between the plates of a parallel plate capacitor as shown. The electric field between the plates will be (JEE Main 2023)

\dfrac{\sigma}{\epsilon_0 (K_1 + K_2)/2}

\dfrac{2\sigma}{\epsilon_0 (K_1 + K_2)}

\dfrac{\sigma (K_1 + K_2)}{\epsilon_0 (K_1 K_2)}

\dfrac{\sigma}{\epsilon_0} \dfrac{2 K_1 K_2}{K_1 + K_2}

The correct solution involves applying the fundamental concept to derive the final value step by step...

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Q.Two thin dielectric slabs of dielectric constant K1K_1K1​ and K2K_2K2​ are inserted between the plates of a parallel plate capacitor as shown. The electric field between the plates will be (JEE Main 2023)