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The capacitance of a parallel-plate capacitor with vacuum is 5 μF. If a dielectric slab with K = 1.5 is inserted between the plates, the capacitance will be equal to ______.

Asked in GSEB Board August 2020 · Dielectric slab

• (1) 0.75 μF
• (2) 7.5 μF
• (3) 3.33 μF
• (4) 0.33 μF

Step-by-step solution

Why the other options are wrong

• (1) 0.75 is 7.5 with the decimal point moved one place: a place-value slip in 5×1.5.
• (3) This divides by K: 5/(1.5) = 3.33 μF. A dielectric weakens the field for a given charge, which lowers V and so raises C by the factor K.
• (4) No single slip gives 0.33 μF (it is 5/(15): dividing by K and also misplacing its decimal point). Sanity check: a dielectric always increases C, so the answer must exceed 5 μF.

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