Hey everyone, My professor just posted this review question: “A parallel-plate c

Question

Hey everyone,

My professor just posted this review question:

“A parallel-plate capacitor has a slab of glass withdielectric constant K = 5 separating the plates. The capacitor isfirst connected to a 9 V battery and then disconnected. Thepotential difference between the plates is measured by connectingthem to a meter that reads 9 V. The slab of glass is nowpulled out. (a) What does the meter now read? (b) What (if any) hasbeen the change in electric potential energy stored in thecapacitor?”

He wrote that the answers are:
a. 45 V b. the final energy is 5 times theinitial energy stored

Unfortunately, I don’t understand how he got those. Couldsomeone please explain this to me? I will definitely rate high ifyou give a good explanation. Thanks so much!

Explanation / Answer

   a.   For an ideal capacitor,after the battery is removed, charge must remain conserved in givencase.    Charge oncapacitor   Q   =   C* V   =   C *5         Coulomb         -------------(1)    Once the slab is removed, let thecapacitance be C0. Now by the definition of dielectricconstant    k   =   C /C0        C0   =   C/ 5      Hence             Q   =   C0*V0      =>   V0   =   Q/C0               or         V0   =   5* Q / C           Here V0   is newpotential difference across capacitor. Substitute the value of Qfrom equation (1)    V0   =   5* C * V / C   =   5 *V   =   5 *9   =   45 V    a.   Intitial potentialenergy      Ui   =   (1/2)* Q * V             Finalpotentialenergy   Uf   =   (1/2)* Q * V0   =   (1/2) * Q *5 * V    Dividing         Uf   /    Ui   =   5   =>   Uf   =   5* Ui    i.e.   P.E.stored becomes five times on removing the dielectric.    i.e.   P.E.stored becomes five times on removing the dielectric.           

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