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A parallel plate air capacitor has a capacitance C. When it is half filled as shown in the figure with a dielectric of dielectric constant K=5, the percentage increase in the capacitance is ______. The dielectric slab of thickness d/2 lies against one plate, the remaining thickness d/2 being air, d being the plate separation.

Asked in JEE Main 2nd April 1st Shift 2026 · Dielectric slab

Figure: Dielectric slab
Answer: (2) 66.67

Step-by-step solution

The two layers sit one above the other, so they behave as two capacitors in series, each of thickness d/2.

Dielectric part: C₁=(Kε₀A)/(d/2)=(2Kε₀A)/d=2KC. Air part: C₂=(ε₀A)/(d/2)=2C.

C'=(C₁C₂)/(C₁+C₂)=(2KC·2C)/(2KC+2C)=(2K)/(K+1)C.

With K=5: C'=(10)/6C=5/3C.

Percentage increase =(C'-C)/C×100=2/3×100=66.67%.

Why the other options are wrong

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