A parallel plate capacitor of plate area A and plate separation d is charged to a potential V and then battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q E and If denote respectively, the magnitude of charge of each plate, the electric field between the plates (after$$ the slab is $$inserted) and work done on the system in question in the process of inserting the slab, then

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Infinity Learn · Accepted Answer

$$C0=$$ε$$0Ad$$ and $$Q0=C0V=$$ε$$0AVd$$                   ...$$(1)When$$ the battery is disconnected, the charge remains the same, i.e.,$$Q=$$ε$$0AV/d$$. Electric field in the dielectric between plates of capacitor                                    $$E=Eair$$ $$K=1KVd$$                       ....$$(2)After$$ the insertion of dielectric, the new capacity C' becomes                              C′$$=K$$ε$$0Ad=KCNow$$,   V′$$=QC$$′$$=QKC=VKWork$$ done on the $$systemW=12CV2$$−$$12C$$′V′$$2=12CV2$$−$$12KCVK2=12CV21$$−$$1K=12$$ε$$0AdV21$$−$$1K$$

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