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

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2. Electric Potential and Capacitance

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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

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

B

${q}_{{b}}={q}_{{f}}\left(1-\frac{1}{\sqrt{{k}}}\right)$

C

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

D

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

(JEE MAIN-2021)

Solution

When a dielectric is inserted in a capacitor

Due to free charge $\vec{E}=\vec{E}_{0}$ only

After dielectric $E^{\prime}=\frac{E_{0}}{k}$

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

Standard 12

Physics

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