Electrostatics Ques 7
7 The parallel combination of two air filled parallel plate capacitors of capacitance $C$ and $n C$ is connected to a battery of voltage, $V$. When the capacitors are fully charged, the battery is removed and after that 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
(Main 2019, 9 April II)
(a) $\frac{(n+1) V}{(K+n)}$
(b) $\frac{n V}{K+n}$
(c) $V$
(d) $\frac{V}{K+n}$
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Answer:
Correct Answer: 7.( a )
Solution:
When battery is removed and a dielectric slab is placed between two plates of first capacitor, then charge on the system remains same. Now, equivalent capacitance after insertion of dielectric is
If potential value after insertion of dielectric is $V^{\prime}$, then charge on system is
$ Q^{\prime}=C_{e q} V^{\prime}=(n+K) C V^{\prime} $
As $Q=Q^{\prime}$, we have
$ \begin{array}{rlrl} C(1+n) V & =(n+K) C V^{\prime} \\ \therefore \quad V^{\prime} =\frac{(1+n) V}{(n+K)} \end{array} $
BETA
