Question

The parallel combination of two air filled parallel plate capacitors of capacitance $C$ and $nC$ is connected to a battery of voltage $V$. When the capacitors are fully charged, the battery is removed and after a dielectric material of dielectric constant $K$ is placed between the two plates of the first capacitor. The new potential difference of the combined system is?
(A) $\dfrac{V}{{\left( {K + n} \right)}}$
(B) $V$
(C) $\dfrac{{\left( {n + 1} \right)V}}{{\left( {K + n} \right)}}$
(D) $\dfrac{{nV}}{{\left( {K + n} \right)}}$

Answer 1

For parallel combination of capacitors, the equivalent or effective capacitance is the sum of all the capacitance of the parallel capacitors. When the battery of a capacitor is removed, the charge is kept constant under all conditions (as long as it is not discharged)
Formula used: In this solution we will be using the following formulae;
$Q = CV$ where $Q$ is the total charge stored in a capacitor, $C$ is the capacitance of the capacitor, and $V$ is the voltage across the capacitor.
${C_{eq}} = {C_1} + {C_2}$ where ${C_{eq}}$ is the equivalent or effective capacitance of two capacitors in a parallel combination, ${C_1}$ and ${C_2}$ are the capacitances of the individual capacitors in the combination.

Complete Step-by-Step Solution:
When the capacitors are charged, the voltage across them is $V$.
The charge on a capacitor is given as
$Q = CV$ where $Q$ is the total charge stored in a capacitor, $C$ is the capacitance of the capacitor, and $V$ is the voltage across the capacitor.
Hence, the charge on each capacitor after charge is
$\Rightarrow$ ${Q_1} = CV$ and
$\Rightarrow$ ${Q_2} = nCV$
Then the total charge in the system would be
$\Rightarrow$ $Q = {Q_1} + {Q_2} = CV + nCV = \left( {n + 1} \right)CV$
Now, a dielectric of dielectric constant $K$ is placed between the plates of the first capacitor, hence, the capacitance becomes
${C_1} = KC$
But the capacitance of the second remains ${C_2} = nC$
For capacitors in parallel, the effective capacitance is given as
$\Rightarrow$ ${C_{eq}} = {C_1} + {C_2}$ where ${C_{eq}}$ is the equivalent or effective capacitance of two capacitors in a parallel combination, ${C_1}$ and ${C_2}$ are the capacitances of the individual capacitors in the combination.
Hence,
$\Rightarrow$ ${C_{eq}} = KC + nC = C\left( {K + n} \right)$
Then the new potential difference would be given as
$\Rightarrow$ ${V_n} = \dfrac{Q}{{{C_{eq}}}} = \dfrac{{\left( {n + 1} \right)CV}}{{C\left( {K + n} \right)}}$
By cancelling common terms, we have
$\Rightarrow$ ${V_n} = \dfrac{{\left( {n + 1} \right)V}}{{\left( {K + n} \right)}}$

Hence, the correct option is C.

Note: For understanding, the formula ${C_{eq}} = {C_1} + {C_2}$ can be proven from the knowledge that the total charge in the system is the sum of the charges in the two capacitors. Hence, since $Q = CV$
We have that
$\Rightarrow$ $Q = {Q_1} + {Q_2} = {C_1}V + {C_2}V$.
$\Rightarrow {C_{eq}}V = {C_1}V + {C_2}V$
It remains $V$ because potential across parallel components are equal, hence,
${C_{eq}} = {C_1} + {C_2}$