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Capacitor question

2019 · 9 Apr · Shift 2 · Q46
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Capacitor question

2019 · 9 Apr · Shift 2 · Q46

JEE MainPhysicsCapacitorMCQ+4 / −1
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 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 :-
  1. A
    V
  2. B
    VK+n{V \over {K + n}}K+nV​
  3. C
    (n+1)VK+n{{(n+1)V} \over {K + n}}K+n(n+1)V​
  4. D
    nVK+n{{nV} \over {K + n}}K+nnV​
View written solutionFree

Correct answer: C

  1. Initial combination and charging

Two capacitors are connected in parallel:

  • First capacitor: C1=CC_1 = CC1​=C
  • Second capacitor: C2=nCC_2 = nCC2​=nC

So the initial equivalent capacitance is Ceq, initial=C+nC=(n+1)C.C_{\text{eq, initial}} = C + nC = (n+1)C.Ceq, initial​=C+nC=(n+1)C.

They are connected to a battery of voltage VVV, so the total charge stored in the parallel combination is Qtotal=Ceq, initial V=(n+1)CV.Q_{\text{total}} = C_{\text{eq, initial}}\,V = (n+1)CV.Qtotal​=Ceq, initial​V=(n+1)CV.

  1. Battery removed

After the battery is removed, the capacitor combination is isolated. Therefore, the total charge of the system remains conserved.

So, Qtotal=(n+1)CV.Q_{\text{total}} = (n+1)CV.Qtotal​=(n+1)CV.

  1. Dielectric inserted in the first capacitor

A dielectric of dielectric constant KKK is inserted into the first capacitor.

Hence its new capacitance becomes C1′=KC.C_1' = KC.C1′​=KC.

The second capacitor remains unchanged: C2′=nC.C_2' = nC.C2′​=nC.

Since they are still in parallel, the new equivalent capacitance is Ceq, new=KC+nC=(K+n)C.C_{\text{eq, new}} = KC + nC = (K+n)C.Ceq, new​=KC+nC=(K+n)C.

  1. Find the new potential difference

Because the system is isolated, Qtotal=Ceq, new V′.Q_{\text{total}} = C_{\text{eq, new}}\,V'.Qtotal​=Ceq, new​V′.

Thus,

= \frac{(n+1)CV}{(K+n)C}.$$ Cancelling $C$, $$V' = \frac{(n+1)V}{K+n}.$$ 5. **Match with options** This corresponds to: $$\boxed{\frac{(n+1)V}{K+n}}$$ which is **Option C**. 6. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** They match.
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