Saved Bookmarks
| 1. |
A parallel plate capacitor of capacitance 40 muF is charged to a potential difference of 250 V and then the battery is disconnected. Separationbetween the paltes of the capacitoris 5 mm. A dielectric slab of dielectric constant 5 and thickness 3 mm is introducedin -between the plates of capacitor. Calcualte final charge and potential difference between the plates. Also calculate energy loss due to the introduction of dielectric slab. |
|
Answer» Solution :Given, initial capacitance of the parallel plate CAPACITOR, `C_(0)=(epsilon_(0)A)/(d)=40muF` Initial potential difference between the plates of capacitor, `V_(0)=250V` Separation between the plates of capacitor, `d=5 mm = 5xx10^(-3)m` Initial charge stored in capacitor, `Q_(0)=C_(0)V_(0)=40xx10^(-6)xx250` `Q_(0)=10^(4)muC` On inserting the DIELECTRIC of thickness t and dielectric constant K, Now the new capacitance will be: `C.=(epsilon_(0)A)/((d-t+(t)/(K)))` `C.=(epsilon_(0)A)/((d-3+(3)/(5))xx10^(-3))` `=(epsilon_(0)A)/((5-3+(3)/(5))xx10^(-3))` `=(epsilon_(0)A)/((13)/(5)xx10^(-3))` We know that, `(epsilon_(0)A)/(5xx10^(-3)) =40muF` `therefore C.=(((epsilon_(0)A)/(5xx10^(-3)))5xx10^(-3))/((13)/(5)xx10^(-3))` `rArr C.=((epsilon_(0)A)/(5xx10^(-3)))xx(5xx10^(-3)xx5)/(13xx10^(-3))` `=(25)/(13)xx40 muF` `C.=76.92muF` Let the new potential difference be V. across plates. New charge on capacitor will be same as before. `therefore V.C.=Q_(0)` `V.=(C_(0)V_(0))/(C.)` `V.=(40xx10^(-6))/(76.92xx10^(-6))xx250` `V.=130V` Initially, energy stored in the capacitor is: `U=(1)/(2)CV^(2)` `=(1)/(2)xx40xx10^(-6)xx(250)^(2)=1.25J` Finally, energy stored in the capacitor is : `U.=(1)/(2)C.(V.)^(2)` `=(1)/(2)xx76.92xx(130)^(2)=0.65J` Energy loss due to introduction of dielectric slab `=U-U.=1.25-0.65=0.60J` |
|