If q_{f} is the free charge on the capacitor | vedclass

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Question

(A) When a dielectric slab is placed between the plates of a capacitor,the net electric field $E$ inside the dielectric is reduced by a factor of $k$

Vedclass · Accepted Answer

$q_{b} = q_{f} \left(1 - \frac{1}{k}\right)$

Vedclass · Answer

(A) When a dielectric slab is placed between the plates of a capacitor,the net electric field $E$ inside the dielectric is reduced by a factor of $k$ compared to the initial electric field $E_{0}$ in vacuum.The net electric field is given by $E = E_{0} - E_{b}$,where $E_{b}$ is the electric field due to the induced (bound) charges.Since $E = \frac{E_{0}}{k}$,we have $\frac{E_{0}}{k} = E_{0} - E_{b}$.This implies $E_{b} = E_{0} - \frac{E_{0}}{k} = E_{0} \left(1 - \frac{1}{k}\right)$.Since the electric field is proportional to the surface charge density $(E = \frac{\sigma}{\epsilon_{0}})$,the bound charge $q_{b}$ is related to the free charge $q_{f}$ by the same factor:$q_{b} = q_{f} \left(1 - \frac{1}{k}\right)$.

If $q_{f}$ is the free charge on the capacitor plates and $q_{b}$ is the bound charge on the dielectric slab of dielectric constant $k$ placed between the capacitor plates,then the bound charge $q_{b}$ can be expressed as:

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