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2. Electric Potential and Capacitance
hard
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
