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JEE Main 20199 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
9 April
Shift
Evening Shift
Subject
Physics
Chapter
Capacitance

JEE Main 2019 · 9 April · Evening Shift · Physics · Capacitance

The parallel combination of two air filled parallel plate...

The parallel combination of two air filled parallel plate capacitors of capacitance CC and nCnC is connected to a battery of voltage, V.V . When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant KK is placed the two plates of the first capacitor. The new potential difference of the combined system is:The parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to a batteryThe parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to a batteryThe parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to a battery

Options

  1. A\dfrac{\left(n + 1\right)V}{\left(K + n\right)}
  2. BnVK+n\dfrac{nV}{K + n}
  3. CVK+n\dfrac{V}{K + n}
  4. DVV

Correct answer

A. \dfrac{\left(n + 1\right)V}{\left(K + n\right)}

Step-by-step solution

i)
The parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to a battery
ii) Battery disconnected after full charge,
Total charge of the system,
=CV+nCV= CV + nCV
=(n+1)CV= \left(n + 1\right)CV
The parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to a battery
iii) Inserting dielectric of constant KK
The parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to a battery
Common potential, VC=totalchargetotalcapacitance{V}_{C} = \dfrac{totalcharge}{totalcapacitance}
=(n+1)CVKC+nC=(n+1)VK+n= \dfrac{\left(n + 1\right)CV}{KC + nC} = \dfrac{\left(n + 1\right)V}{K + n}
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