Question

For making a parallel plate capacitor, two plates of copper, a sheet of mica (thickness = 0.1mm , K=5.4) , a sheet of glass (thickness = 1.0 cm, K=2) are available. To obtain the largest capacitance, which sheet should you place between the copper plates.
A) Mica
B) Glass
C) Paraffin
D) None of these

Answer 1

Hint: Parallel plate capacitor is a device which is used to store the charge. In this device we use two parallel plate sheets and between these two plates we use material called dielectric. Measurement unit of a capacitor is called capacitance.
Formula to calculate capacitance is: \[C = \dfrac{{k{ \in _o}A}}{d}\]

Complete step by step solution:
We know that if the same charge is flowing through the capacitors then they are in series and the charge is different then, they are in parallel.
Also we know that if the thickness is different and area is same then they are in series. As it is given in this case.
Firstly we will calculate the capacitance using dielectric as mica:
Given for mica:
K=5.4 and d=0.1 mm
So apply capacitance equation for mica:
$
  {c_m} = \dfrac{{k{ \in _o}A}}{d} \\
  {c_m} = \dfrac{{5.4 \times { \in _o}A}}{{{{10}^{ - 4}}}} \\
  {c_m} = 5.4 \times {10^4}{ \in _o}A $
Now we will calculate for glass as dielectric:
K=2 and d=1 cm.
Apply capacitance equation for glass:
$
  {c_g} = \dfrac{{k{ \in _o}A}}{d} \\
  {c_g} = \dfrac{{2 \times { \in _o}A}}{{{{10}^{ - 2}}}} \\
  {c_g} = 2 \times {10^2}{ \in _o}A $
So by the above two observations we can say capacitance of capacitor will be more if we use mica as dielectric material.
${c_m} > {c_g}$

Hence the correct answer will be option A.

Note: Capacitance is the ratio of change in electric charge of a system to the corresponding change in its electric potential.
Mathematically:
$
  Q = CV \\
  C = \dfrac{Q}{V} $
where Q is charge stored, V Is potential and C is capacitance