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(A) The electric field $(E)$ inside a dielectric medium in a capacitor connected to a battery is given by $E = \frac{V}{d_{eff}}$,where $V$ is the pot

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(A) The electric field $(E)$ inside a dielectric medium in a capacitor connected to a battery is given by $E = \frac{V}{d_{eff}}$,where $V$ is the potential difference across the plates and $d_{eff}$ is the effective distance. For a dielectric slab of thickness $t$ and dielectric constant $K$,the electric field is $E = \frac{V}{K \cdot d_{total}}$. More simply,for a given surface charge density $\sigma$ on the plates,the electric field in a medium with dielectric constant $K$ is $E = \frac{\sigma}{K \epsilon_0}$. Since the capacitor is connected to a battery,the potential difference $V$ across the plates remains constant. In the first region $(0 In the second region $(d Comparing the two,we see that $E_1 = 2 E_2$. Therefore,the electric field in the first region is twice the electric field in the second region,and both are constant within their respective regions. This corresponds to the graph where the electric field is higher for $0 < x < d$ and lower for $d < x < 3d$.

$A$ parallel plate capacitor has two layers of dielectric as shown in the figure. This capacitor is connected across a battery. The graph between electric field $(E)$ and distance $(x)$ from the left plate will be:

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