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A parallel plate capacitor with air between the plates has a capacitance of 1.0 pF. If the distance between the plates is made doubled and space between them is filled with dielectric substance, the capacitance becomes 2.0 pF. Then the value of dielectric constant of dielectric substance is ________.

Asked in GUJCET 2026 · Dielectric slab

• (1) 1.5
• (2) 2.0
• (3) 3.0
• (4) 4.0

Step-by-step solution

Why the other options are wrong

• (1) No single slip gives 1.5. Sanity check: doubling the gap halves C, so C'=K/2×1.0 pF; with K=1.5 the capacitor would end at 0.75 pF, and reaching 2.0 pF needs K=4.
• (2) Ignores the doubled gap and reads K straight off the ratio (C')/C=(2.0)/(1.0)=2. Doubling d on its own halves the capacitance, so the dielectric has to make up a factor of 4, not 2.
• (3) With K=3 the new capacitance would be 3/2×1.0=1.5 pF, not 2.0 pF.

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