9. 01 points | Previcus Answers OSUniPhys1 25.4.WA.039 My Notes Ask Your Teacher

Question

9. 01 points | Previcus Answers OSUniPhys1 25.4.WA.039 My Notes Ask Your Teacher Two plates, each of area 5.20 cm2, are 2.59 mm apart in a parallel plate capacitor. The space between the plates is filled with two different dielectric materials. The left half of the capacitor is filled with a material whose dielectric constant K, 5.80. The right half of the capacitor is filled with a material of unknown dielectric value K2. The capacitance of this device is 8.25 pF. Find the value of 3.86 The arrangement is similar to two capacitors in parallel. What is the equivalent capacitance in this case? Additional Materials eBook

Explanation / Answer

Capacitance of air filled capacitor is given by:

C = e0*A/d

If a dielectric material (k) is inserted then capacitance is given by:

C1 = k*e0*A/d

k = dielectric constant of that material

Now given that

A = 5.20 cm^2

d = 2.59 mm

Now in this scenario when two dielectric materials are inserted they will be connected in parallel circuit. In parallel, equivalent capacitance is given by:

Ceq = C1 + C2

C1 = k1*eo*(A/2)/d = k1*e0*A/2d

C2 = k2*e0*(A/2)/d = k2*e0*A/2d

Ceq = 8.25 pF = 8.25*19^-12 F

Using these values

Ceq = k1*e0*A/2d + k2*e0*A/2d

Ceq = (k1 + k2)*e0*A/(2d)

k2 = Ceq*2*d/(e0*A) - k1

k2 = 8.25*10^-12*2*2.59*10^-3/(8.85*10^-12*5.20*10^-4) - 5.80

k2 = 3.49 = dielectric constant of second material

Please Upvote.

Comment below if you have any doubt.

Related Questions

Navigate

• Browse (All)
• Browse 9
• Subjects

---

---

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