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JEE MAIN 2021
25-07-21 S2
Question
If $\mathrm{g}_{\mathrm{f}}$ is the free charge on the capacitor plates and $\mathrm{g}_{\mathrm{b}}$ is the bound charge on the dielectric slab of dielectric constant k placed between the capacitor plates, then bound charge $\mathrm{g}_{\mathrm{b}}$ can be expressed as
Select the correct option:
A
$q_b=q_f\left(1-\frac{1}{\sqrt{k}}\right)$
B
$q_b=q_f\left(1-\frac{1}{k}\right)$
C
$q_b=q_f\left(1+\frac{1}{\sqrt{k}}\right)$
D
$q_b=q_f\left(1+\frac{1}{k}\right)$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
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Physics
Easy
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