Question 98 of 180

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 :

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${q}_{b}={q}_{f}\left(1+\dfrac{1}{k}\right)$

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

When a dielectric is inserted in a capacitor,
${q}_{b}={q}_{i}\left(1-\dfrac{1}{k}\right)$