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If \mathrm{q}_{\mathrm{f}} is the free charge on the capacitor plates and \mathrm{q}_{\mathrm{b}} is the bound charge on the dielectric slab of dielectric constant \mathrm{k}placed between the capacitor plates, then the bound charge \mathrm{q}_{\mathrm{b}} can be expressed as :
 
Option: 1 \mathrm{q}_{\mathrm{b}}=\mathrm{q}_{\mathrm{f}}\left(1-\frac{1}{\sqrt{\mathrm{k}}}\right)
Option: 2 \mathrm{q}_{\mathrm{b}}=\mathrm{q}_{\mathrm{f}}\left(1-\frac{1}{\mathrm{k}}\right)
Option: 3 \mathrm{q}_{\mathrm{b}}=\mathrm{q}_{\mathrm{f}}\left(1+\frac{1}{\sqrt{\mathrm{k}}}\right)
Option: 4 \mathrm{q}_{\mathrm{b}}=\mathrm{q}_{\mathrm{f}}\left(1+\frac{1}{\mathrm{k}}\right)

Answers (1)

best_answer


E_{net}\rightarrow Net electric field inside dielectric slab
E_{net}= \frac{E_{ext}}{k}= E_{ext}-E_{ind}
E_{induced}= E_{ext}\left ( 1-\frac{1}{k} \right )
\frac{q_{b}}{A\epsilon _{0}}= \frac{q _{f}}{A\epsilon _{0}}\left ( 1-\frac{1}{k} \right )
q_{b}= q _{f}\left ( 1-\frac{1}{k} \right )
The correct option is (2)

Posted by

vishal kumar

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