EAMCET Engineering 2003 Solved Paper

Capacitance of parallel plate capacitor,
C0=‌

ε0A

d

where, A= area of the plates,
d= separation between the plates
Charge stored in the capacitor,
Q=C0V0
When battery is disconnected, then charge remains same.
So, energy, E1=‌

1

2

‌

Q2

C

C= capacitance when plate separation is doubled
So, ‌‌C1=‌

C0

2

E1=‌

1

2

‌

Q2

C0∕2

=‌

Q2

C0

=‌

C02V02

C0

=C0V02
When battery is connected, then
Energy, E2=‌

1

2

CV02
where, E2=‌

1

2

‌

C0

2

V02=‌

1

4

(C0V02)
‌∴‌‌‌

E1

E2

=‌

C0V02

1 4 C0V02

=‌

1

4

‌E1:E2=4:1