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(B) The electric field in a dielectric medium is given by $E = \frac{\sigma}{\varepsilon_0 K} = \frac{Q}{A \varepsilon_0 K}$.Let $E_1, E_2, E_3$ be th

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$\frac{3}{K_2} = \frac{2}{K_1} + \frac{1}{K_3}$

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(B) The electric field in a dielectric medium is given by $E = \frac{\sigma}{\varepsilon_0 K} = \frac{Q}{A \varepsilon_0 K}$.Let $E_1, E_2, E_3$ be the electric fields in the regions with dielectric constants $K_1, K_2, K_3$ respectively.From the graph,the change in electric field $x = E_1 - E_2 = \frac{Q}{A \varepsilon_0 K_1} - \frac{Q}{A \varepsilon_0 K_2}$.Similarly,the change in electric field $y = E_2 - E_3 = \frac{Q}{A \varepsilon_0 K_2} - \frac{Q}{A \varepsilon_0 K_3}$.Given the condition $2x = y$,we substitute the expressions:$2 \left( \frac{Q}{A \varepsilon_0 K_1} - \frac{Q}{A \varepsilon_0 K_2} \right) = \frac{Q}{A \varepsilon_0 K_2} - \frac{Q}{A \varepsilon_0 K_3}$.Dividing both sides by $\frac{Q}{A \varepsilon_0}$,we get:$2 \left( \frac{1}{K_1} - \frac{1}{K_2} \right) = \frac{1}{K_2} - \frac{1}{K_3}$.Rearranging the terms:$\frac{2}{K_1} - \frac{2}{K_2} = \frac{1}{K_2} - \frac{1}{K_3}$.$\frac{2}{K_1} + \frac{1}{K_3} = \frac{1}{K_2} + \frac{2}{K_2} = \frac{3}{K_2}$.Thus,the relation is $\frac{3}{K_2} = \frac{2}{K_1} + \frac{1}{K_3}$.

The variation of the electric field between the plates of a capacitor is shown in the figure. Find the relation between $K_1$,$K_2$,and $K_3$,if $2x = y$.

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