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) $\frac{V}{K+n}$
Show Answer
Answer:
Correct Answer: 7.( a )
Solution:
[Image]
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
[Image]
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
