Electrostatic Potential and Capacitance Part 2

As per the arrangement given, distance between the capacitor plates are a and b and a≠b using the diagram we can write b=5−a−1=(4−a) in mm
as we know capacitance of capacitor C=‌

ε0A

d

and in series arrangement
‌

1

C‌eq ‌

=‌

1

C1

+‌

1

C2

‌

1

C‌eq ‌

=‌

a

ε0A

+‌

4−a

ε0A

=‌

4(‌ in mm ‌)

ε0A

or Ceq=‌

ε0A

4(mm)

Given A=200cm2
Ceq=‌

ε0×200×10−4

4×10−3

=ε050×10−1
or Ceq=5ε0 farad therefore n=5