VITEEE 2016 Paper

When a dielectric slab of dielectric constant
K =

3

2

is inserted between the plates of C2 , its new capacitance (C2′) become
C2′ =

3

2

C
Equivalent capacitance of C2′ and C3 is
Ceq = C2′+C3 =

3

2

C + C =

5C

2

Now, Ceq and C1 are in series. Therefore, their equivalent capacitance is
Ceq =

Ceq×C1

Ceq+C1

=

2 5C 2 ×C

5C 2 +C

=

5C2

7C

=

5C

7