AP EAPCET 20-Aug-2021 Shift 1 Paper

Given,
Capacitance of parallel plate capacitor, C0=60µF =60×10−6F
Separation between plates, d=6‌mm =6×10−3m
Electric potential, V0=250V
Dielectric constant of slab, K=5
and thickness of slab, t=3‌mm =3×10−3m
Since, C=

q

V

=

ε0A

d

=

ε0A

d−t(1− 1 K )

where, q is charge stored, V is potential,
ε0 is free space permittivity =8.854×10−12Nm2C−2 and A is plate area
∴q0=C0V0 =60×10−6×250 =15×10−3C
and new capacitance, C′=

ε0A

[6−3(1− 1 5 )]×10−3

...(i)
and ε0A=C0d
⇒ε0A=60×10−6×6×10−3 =360×10−9
Put this value in Eq. (i), we get
C′=

360×10−9

3.6×10−3

=10−4F
Final potential stored in capacitor =

q0

C′

=15×10−310−4=150V
Hence, voltage difference =250−150=100V