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