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Capacitor question

2015 · Q115
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Capacitor question

2015 · Q115

NEETPhysicsCapacitorMCQ+4 / −1
A parallel plate air capacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dieletric constant K, which can just fill the air gap of the capacitor, is now inserted in it . Which of the following is incorrect ?
  1. A
    The change in energy stored is 12CV2(1K−1){1 \over 2}C{V^2}\left( {{1 \over K} - 1} \right)21​CV2(K1​−1)
  2. B
    The charge on the capacitor is not conserved.
  3. C
    The potential difference between the plates decreases K times.
  4. D
    The energy stored in the capaciotor decreases K times.
View written solutionFree

Correct answer: B

AIPMT 2015 Cancelled Paper Physics - Capacitor Question 32 English Explanation

q = CV ⇒\Rightarrow⇒ V = q/C

Due to dielectric insertion, new capacitance C2 = CK

Initial energy stored in capacitor, U1=q22C{U_1} = {{{q^2}} \over {2C}}U1​=2Cq2​

Final energy stored in capacitor, U2=q22KC{U_2} = {{{q^2}} \over {2KC}}U2​=2KCq2​

Change in energy stored, Δ\Delta ΔU = U2 – U1

ΔU=q22C(1K−1)=12CV2(1K−1)\Delta U = {{{q^2}} \over {2C}}\left( {{1 \over K} - 1} \right) = {1 \over 2}C{V^2}\left( {{1 \over K} - 1} \right)ΔU=2Cq2​(K1​−1)=21​CV2(K1​−1)

New potential difference between plates

V′=qCK=VKV' = {q \over {CK}} = {V \over K}V′=CKq​=KV​

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