The figure below shows a parallel plate capacitor of plate area A = 120 cm 2 and

Question

The figure below shows a parallel plate capacitor of plate area A = 120 cm2 and plate separation d = 1.35 cm. A potential difference of V0 = 90.0 V is applied between the plates. Suppose that the battery remains connected while the dielectric slab of thickness b = 0.780 cm and dielectric constant = 3.20 is being introduced. Calculate the following values.

(a) the capacitance: 13.05 pF

(b) the charge on the capacitor plates: 1.17 nC

(c) the electric field in the gap: ????? N/C

(d) the electric field in the slab, after the slab is in place: 3442 N/C

What is (c)???

Explanation / Answer

given that

A = 120cm^2 = 120*10^(-4) m^2

d = 1.35 cm = 0.0135 m

b = 0.780 cm = 0.78*10^(-2) m

k = 3.20

Initial C

Parallel plate cap

C =e0*k*(A/d) F

e0 is vacuum permittivity, 8.854*10^(-12) F/m

k is dielectric constant or relative permittivity of the material (vacuum = 1)

A and d are area of plate and separation .

a)
initial C0 = (8.854*10^(-12))*(120*10^(-4)) / (0.0135)

C0 = 7.87*10^(-12)F

final C, we can see that here two capacitors in series

C1 is an air capacitor with spacing 1.35 – 0.78 = 0.57 cm

C2 is a cap with spacing 0.78 and a dielectric

C1 = (8.854*10^(-12))*(120*10^(-4)) / (0.0057)

C1 = 18.64*10^(-12) F

C2 = (8.854*10^(-12))*(120*10^(-4))*(3.20) / (0.0078)

C2 = 43.58*10^(-12) F

final C = C1*C2 / (C1+C2)

final C = 18.64*10^(-12) * 43.58*10^(-12) / 62.22*10^(-12)

final C = 13*10^(-12) F = 13 pF

b)
Q = C*V

Q = 13*10^(-12)*90

Q = 1.17*10^(-12) C = 1.17 pC

part(c)

we know that

E = V / d

E = 90 / 1.35*10^(-2)

E = 6666 N /C

part(d)

E = 90 / 0.78*10^(-2)

E = 11538 N / C

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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