Answer

Verified

95.1k+ views

**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,

\[{C_{eq}} = {C_1} + {C_2}\]

Recently Updated Pages

Write a composition in approximately 450 500 words class 10 english JEE_Main

Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main

Write an article on the need and importance of sports class 10 english JEE_Main

Name the scale on which the destructive energy of an class 11 physics JEE_Main

Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main

Choose the one which best expresses the meaning of class 9 english JEE_Main