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NEET 2015
Exam
NEET
Year
2015
Subject
Physics
Chapter
Capacitance

NEET 2015 · Physics · Capacitance

A parallel plate air capacitor of capacitance C is connected to a...

A parallel plate air capacitor of capacitance CC is connected to a cell of emf VV and then disconnected from it. A dielectric slab of dielectric constant KK, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A

Options

  1. AThe potential difference between the plates decreases KK times.
  2. BThe energy stored in the capacitor decreases KK times.
  3. CThe change in energy stored is 12CV2(1K−1)\dfrac{1}{2}C{V}^{2}\left(\dfrac{1}{K} - 1\right).
  4. DThe charge on the capacitor is not conserved.

Correct answer

D. The charge on the capacitor is not conserved.

Step-by-step solution

A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A
For air K=1K = 1,
C1=C{C}_{1} = C
q=C1V…(i)q = {C}_{1}V \ldots \left(i\right).
The charge remains constant.
U1=q22C{U}_{1} = \dfrac{{q}^{2}}{2C}
U2=q22CK{U}_{2} = \dfrac{{q}^{2}}{2CK}
Change in energy stored in the capacitor,
U=U2−U1U = {U}_{2} - {U}_{1}
=q22C(1K−1)=12CV2(1K−1)= \dfrac{{q}^{2}}{2C}\left(\dfrac{1}{K} - 1\right) = \dfrac{1}{2}C{V}^{2}\left(\dfrac{1}{K} - 1\right)
V′=qCK⇒V′=VKV ' = \dfrac{q}{CK} ⇒ V ' = \dfrac{V}{K}
Here,1- Potential difference between the plates decreases by KK times.
2- The change in energy stored ∆U=12CV2(1K−1)∆ U = \dfrac{1}{2}C{V}^{2}\left(\dfrac{1}{K} - 1\right).
3- Charge is conserved.
4- The energy stored in capacitor decrease by K,U2=12q2CKK, {U}_{2} = \dfrac{1}{2}\dfrac{{q}^{2}}{CK}.
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