If q _ f is the free charge on the capacitor plates and q _ b is...

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 bound charge ${q}_{b}$ can be expressed as :

Options

A ${q}_{b} = {q}_{f}\left(1 - \dfrac{1}{\sqrt{k}}\right)$
B ${q}_{b} = {q}_{i}\left(1 - \dfrac{1}{k}\right)$
C ${q}_{b} = {q}_{f}\left(1 + \dfrac{1}{\sqrt{k}}\right)$
D ${q}_{b} = {q}_{f}\left(1 + \dfrac{1}{k}\right)$

Correct answer

B. ${q}_{b} = {q}_{i}\left(1 - \dfrac{1}{k}\right)$

Step-by-step solution

If q _ f is the free charge on the capacitor plates and q _ b is the bound charge on the dielectric slab of dielectric

When a dielectric is inserted in a capacitor,

${q}_{b} = {q}_{i}\left(1 - \dfrac{1}{k}\right)$

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