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(D) The negative plate is at $x = 0$ and the positive plate is at $x = 3d$. The electric field lines are always directed from the positive plate to th

Vedclass · Accepted Answer

Both $(b)$ and $(c)$.

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(D) The negative plate is at $x = 0$ and the positive plate is at $x = 3d$. The electric field lines are always directed from the positive plate to the negative plate. Therefore,the direction of the electric field is from $x = 3d$ towards $x = 0$ (i.e.,in the negative $x$-direction) throughout the region. Thus,the direction remains the same.The magnitude of the electric field in the air gap is $E_{air} = \frac{\sigma}{\varepsilon_0}$ and in the dielectric slab is $E_{dielectric} = \frac{\sigma}{K\varepsilon_0}$. Since $K > 1$,the magnitudes are different.Electric potential $V$ is related to the electric field $E$ by $E = -\frac{dV}{dx}$. Since the electric field is directed towards the negative $x$-axis,$E_x  0$. This means the electric potential $V$ increases continuously as we move from $x = 0$ (negative plate) to $x = 3d$ (positive plate).Therefore,both statements $(b)$ and $(c)$ are correct.

$A$ dielectric slab of thickness $d$ is inserted in a parallel plate capacitor whose negative plate is at $x = 0$ and positive plate is at $x = 3d$. The slab is equidistant from the plates. The capacitor is given some charge. As one goes from $x = 0$ to $x = 3d$:

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