A dielectric slab of dielectric constant K is | vedclass

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(B) When a dielectric slab is placed in an external electric field $E_0$ produced by the capacitor plates,the molecules of the dielectric polarize.Thi

Vedclass · Accepted Answer

$q^{\prime} = -q \left(1 - \frac{1}{K}\right)$

Vedclass · Answer

(B) When a dielectric slab is placed in an external electric field $E_0$ produced by the capacitor plates,the molecules of the dielectric polarize.This polarization creates an induced electric field $E_i$ inside the dielectric,which opposes the external field.The net electric field inside the dielectric is $E = E_0 - E_i = \frac{E_0}{K}$.Since $E_0 = \frac{\sigma}{\epsilon_0} = \frac{q}{A\epsilon_0}$ and $E_i = \frac{\sigma^{\prime}}{\epsilon_0} = \frac{q^{\prime}}{A\epsilon_0}$,we have:$\frac{q}{A\epsilon_0} - \frac{q^{\prime}}{A\epsilon_0} = \frac{q}{K A \epsilon_0}$$q - q^{\prime} = \frac{q}{K}$$q^{\prime} = q - \frac{q}{K} = q \left(1 - \frac{1}{K}\right)$.Since the induced charge on the surface facing the positive plate is negative,the magnitude is $q^{\prime} = -q \left(1 - \frac{1}{K}\right)$.

$A$ dielectric slab of dielectric constant $K$ is placed between the plates of a parallel plate capacitor carrying charge $q$. The induced charge $q^{\prime}$ on the surface of the slab is given by

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