UPSEE 2019 Solved Paper

Initially, a capacitor of capacitance C is connected to a battery of potential V, then the charge stored
Q=

C

V

If battery is removed then stored charge remains constant. i.e.
Q= constant (∵ conservation of charge )
As, new capacitance of the capacitor
C′=εr(

ε0A

d

)
⇒ C′=εrC[∵C=

ε0A

d

]
Which shows that capacitance is increased.
Potential across new capacitor,
V′=

C′

Q

=εr

C

Q

=εrV[∵V=

C

Q

]
So, potential difference across capacitor increased.
Energy stored in new capacitor and old capacitor.
E=

1

2

CV2 and
E′=

1

2

C′V′2=

1

2

(εrC)(εrV)2
⇒E′=

1

2

εr3CV2
Hence, the energy associated with capacitor is increased.