JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2021
- Hard
- MCQ
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
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Answer & Solution
Correct Answer
(A) \({q}_{{b}}={q}_{{f}}\left(1-\frac{1}{{k}}\right)\)
Step-by-step Solution
Detailed explanation
When a dielectric is inserted in a capacitor
Due to free charge \(\vec{E}=\vec{E}_{0}\) only
After dielectric \(E^{\prime}=\frac{E_{0}}{k}\)
\(q_{B}=q_{f}\left(1-\frac{1}{k}\right)\)
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