Question

A parallel plate capacitor with air as a dielectric is charged to a potential “V” using a battery. Removing the battery, the charged capacitor is then connected across an identical uncharged parallel plate capacitor filled with wax of dielectric constant “k” the common potential of both the capacitor is
A) ${\text{V}}$ Volts
B) ${\text{kV}}$ Volts
C) $\left( {{\text{k + 1}}} \right)\;{\text{V}}$ Volts
D) $\dfrac{{\text{V}}}{{{\text{k + 1}}}}$ Volts

Answer 1

The charge of a capacitor is found by calculating the product of capacitance and voltage across the capacitor. The capacitance will increase when we use a dielectric between the plates. The total capacitance varies according to the series and parallel connection of the capacitors.

Complete step by step answer:
Given, a parallel plate capacitor with air as a dielectric is charged to a potential ${\text{V}}$ using a battery.
The expression for the charge is given as,
$Q = CV$
Where $C$ is the capacitance and ${\text{V}}$ is the potential across the capacitor.
The expression for capacitance where the air is the dielectric is given as,
$C = \dfrac{{{\varepsilon _0}A}}{d}$
Where ${\varepsilon _0}$ is the permittivity of the free space, $A$ is the area of one plate and $d$ is the distance between the plates.
Also given an identical uncharged parallel plate capacitor filled with wax of dielectric constant $K$.
The expression for capacitance with a dielectric between the plates is given as,
$C' = K{\varepsilon _0}\dfrac{A}{d}$
Where $K$ is the dielectric constant of the material used. Hence the parallel plate capacitors are identical, the area of each plate and distance between two plates are the same.
From the above two equations, it is clear that $C' = KC$.
We have to find the common potential of two capacitors. Therefore they are connected in parallel. When the capacitors are connected parallel, then the total capacitance is given as,
${C_T} = C + C'$
Substituting the above result we get,
${C_T} = C + KC$
The potential difference is common and the expression is given as,
$V = \dfrac{Q}{{{C_T}}}$
Where $Q$ is the total charge.
Substituting for the variables in the above expression,
$V = \dfrac{{CV}}{{C + KC}} \\\ \Rightarrow V= \dfrac{{CV}}{{C\left( {1 + K} \right)}} \\\ \Rightarrow V= \dfrac{V}{{K + 1}} \\\$

The common potential of the two capacitors is $\dfrac{V}{{K + 1}}$ Volts. Thus, the answer is option D.

Note:
If the external forces are balanced only the body will remain in its state. This is because the external forces will cancel each other to balance. If the external forces are not balanced it will change the state of rest or uniform motion.