Question

The capacitance of a parallel plate condenser is $5\mu F$. When a glass plate is placed between the plates of the condenser, its potential difference reduces to ${\left( {1/8} \right)^{th}}$ of original value. The magnitude of relative dielectric constant of glass is
(A) 2
(B) 6
(C) 7
(D) 8

Answer 1

Hint:
For any parallel plate capacitor having a medium with a dielectric constant $k$ in between its plates, the capacitance is given by $C = \dfrac{{k{\varepsilon _o}A}}{d}$ and the potential difference is given by, $V = \dfrac{Q}{C}$. So from here we can find the potential difference in the two cases and equate them to get the value of the dielectric constant.
To solve this question, we will be using the following formulae:
$\Rightarrow C = \dfrac{{k{\varepsilon _o}A}}{d}$,
where $C$ is the Capacitance of the parallel plate capacitor.
$k$ is the dielectric constant of the medium,
${\varepsilon _o}$ is the permeability is free space
$A$ is the area of the parallel plates
and $d$ is the distance between the parallel plates.
$\Rightarrow V = \dfrac{Q}{C}$
where $V$ is the potential difference between the plates and $Q$ is the charge between the plates.

Complete step by step answer:
For a parallel plate capacitor, the capacitance between the plates is given by the formula,
$\Rightarrow C = \dfrac{{k{\varepsilon _o}A}}{d}$
When the medium in between the plates is air, then the value of the dielectric constant is 1 and the capacitance is given by
$\Rightarrow C = \dfrac{{{\varepsilon _o}A}}{d}$
If the plates are carrying a charge $Q$ then the potential in between the plates is given by
$\Rightarrow V = \dfrac{Q}{C}$
Now substituting the value of $C$ in the equation, we get
$\Rightarrow V = \dfrac{{Qd}}{{{\varepsilon _o}A}}$
This is the potential difference between the plates where there is no dielectric medium present.
Now, according to the question, if a dielectric that is glass is introduced between the plates, the capacitance changes to
$\Rightarrow C' = \dfrac{{k{\varepsilon _o}A}}{d}$
where $k$ is the dielectric constant of glass.
Therefore as the capacitance between the plates change, the value of the potential difference between the plates also changes. So we get,
$\Rightarrow V' = \dfrac{Q}{{C'}}$
So by substituting the value of $C'$
$\Rightarrow V' = \dfrac{{Qd}}{{k{\varepsilon _o}A}}$
According to the question, the potential difference between the plates when the dielectric is placed is ${\left( {1/8} \right)^{th}}$ of the value of the potential difference when no dielectric is present.
Therefore, $V' = \dfrac{1}{8}V$
Now substituting the values of $V$ and $V'$ we get,
$\Rightarrow \dfrac{{Qd}}{{k{\varepsilon _o}A}} = \dfrac{1}{8}\dfrac{{Qd}}{{{\varepsilon _o}A}}$
We can cancel $Q,d,{\varepsilon _o}$ and $A$ from both the sides of the equation we get,
$\Rightarrow \dfrac{1}{k} = \dfrac{1}{8}$
Inverting both the sides of the equation we get
$\Rightarrow k = 8$
Therefore the magnitude of the relative dielectric constant of glass is 8.
So the correct answer will be option (D); 8.

Note:
In the question, we are given the term parallel plate condenser. The condenser is another term that was used earlier to define a capacitor. The properties of a parallel plate capacitor make it suitable to filter the harmonics from an AC supply. It is also used to tune electronics for various purposes.