The figure shows a parallel-plate capacitor with a plate area A = 1.08 cm 2 and

Question

The figure shows a parallel-plate capacitor with a plate area A = 1.08 cm2 and plate separation d = 6.13 mm. The top half of the gap is filled with material of dielectric constant 1 = 13.4; the bottom half is filled with material of dielectric constant 2 = 12.2. What is the capacitance?

Chapter 25, Problem 049 The figure shows a parallel-plate capacitor with a plate area A 1.08 cm2 and plate separation d 6.13 mm. The top half of the gap is filled with material of dielectric constant K1-13.4; the bottom half is filled with material of dielectric constant K2 = 12.2, what is the capacitance? Ky Number Units the tolerance is +/-396

Explanation / Answer

We know that -

C = epcilo0*k*A / d

take upper and lower as two capacitors and C1, C2 as their capacitances.

So, C1 = (8.854x10^-12x13.4x1.08x10^-4) / (3.06x10^-3) = 2.09 x 10^-12 F = 4.18 pF

likewise -

C2 = (8.854x10^-12x12.2x1.08x10^-4) / (3.06x10^-3) = 2.09 x 10^-12 F = 3.80 pF

Now, the two capacitors are in series -

So, Ceq = C1*C2 / (C1+C2) = (4.18*3.80) / (4.18+3.80) = 2.0 pF (This is your answer).

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.