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

Hint: 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,
\[{C_{eq}} = {C_1} + {C_2}\]