AP EAMCET Engineering 2018 Apr 24 Shift 1 Paper

The potential difference before inserting insulator is,
V=

σd

ε0

100=

σd

ε0

The expression of the potential after inserting 2‌mm of insulator is,
V1=E1(2)+E2(d−2)
V1=

σ

ε0k

(2)+

σ

ε0

(d−2)
V1=

σ

ε0

(

2

k

+d−2)
To maintain potential difference, same distance is increasedby 1.6mm. So the expression of extra potential difference is,
V2=

σ

ε0

(1.6)
So,
V1+V2=V
Substitute values.

σ

ε0

(

2

k

+d−2)+

σ

ε0

(1.6)=

σd

ε0

σd

ε0

−

2σ

ε0k

−

σ

ε0

(d−2)=

σ

ε0

(1.6)

2σ

ε0

−

2σ

ε0k

=

σ

ε0

(1.6)
k=5