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A parallel-plate capacitor has plate separation d. A dielectric slab…
Question
A parallel-plate capacitor has plate separation d. A dielectric slab of dielectric constant 4 is inserted parallel to the plates, occupying a thickness 3d/4; the remaining thickness d/4 is vacuum. Let C0 be the capacitance of the same capacitor when completely empty. The new capacitance equals (p/q) C0 in lowest terms. Report p + q.
✓ Verified answer: 23checked by our engine — not a guess
Step-by-step solution
The filled capacitor is two capacitors in series.
The vacuum part has thickness d/4, so its capacitance C_a = eps0*A/(d/4) = 4 C0 (since C0 = eps0*A/d).
The dielectric part has thickness 3d/4 and constant K=4, so C_b = K*eps0*A/(3d/4) = 4*(4/3) C0 = 16/3 C0.
In series: C = C_a*C_b/(C_a+C_b) = (4 * 16/3)/(4 + 16/3) C0 = (64/3)/(28/3) C0 = 64/28 C0 = 16/7 C0.
Lowest terms 16/7, so p+q = 16+7 = 23.
Final answer23
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