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MHT CET 2026 (13 Apr, Shift 2) · Dielectrics

Dielectrics question from MHT CET 2026 (13 Apr, Shift 2)

Q3MHT CET 2026 · 13 APR · SHIFT 2Easy

Dielectrics

A parallel plate capacitor with air between the plate has a capacitance of 1515 pF. The separation between the plates becomes twice and the space between them is filled with a medium of dielectric constant 3.53.5. Then the capacitance becomes x/4x/4 pF.

The value of xx is

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Answer: (a)

  1. C=Kε0AdC = \frac{K\varepsilon_0 A}{d}; new C′=3.5 ε0A2d=3.52CC' = \frac{3.5\,\varepsilon_0 A}{2d} = \frac{3.5}{2}C.
  2. C′=1.75×15=26.25 pF=1054 pFC' = 1.75 \times 15 = 26.25\text{ pF} = \frac{105}{4}\text{ pF}
  3. So the answer is (a) x=105x = 105.
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