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MHT CET 2026 (14 May, Shift 1) · Dielectrics

Dielectrics question from MHT CET 2026 (14 May, Shift 1)

Q13MHT CET 2026 · 14 MAY · SHIFT 1Medium

Dielectrics

A parallel plate capacitor having capacity 1212 pF is charged by a battery of 1010 V between its plates. The battery is now disconnected and a slab of dielectric constant 66 is kept between the plates. The work done by the capacitor on the slab is

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Answer: (c)

  1. Battery disconnected, so QQ is fixed and U′=UKU' = \frac{U}{K}.
  2. U=12CV2=12×12×10−12×100=6×10−10U = \frac{1}{2}CV^2 = \frac{1}{2} \times 12 \times 10^{-12} \times 100 = 6 \times 10^{-10} J.
  3. U′=6×10−106=1×10−10U' = \frac{6 \times 10^{-10}}{6} = 1 \times 10^{-10} J.
  4. Work done on the slab =U−U′=5×10−10= U - U' = 5 \times 10^{-10} J.
  5. So the answer is (c).
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