A parallel place capacitor has an equivalent plate area of A = 1.9 m^2 and is ch

Question

A parallel place capacitor has an equivalent plate area of A = 1.9 m^2 and is charged to 12 volts. The dielectric has a permittivity of r = 6 and a plate separation of d = 0.0001 m. Including proper units, calculate the capacitance C from C = r 0 A/d the total charge Q stored on the capacitor; the surface charge density on the plates, s; the electric field E between the plates of the capacitor; the electric displacement D in the dielectric. Compare D to the surface charge density s. the electric polarization P in the dielectric; the electric susceptibility E of the dielectric; the total energy stored in the capacitor. What fraction of the energy is stored in the electric field, and what percentage is stored in the dielectric?

Explanation / Answer

Only 4 subparts at a time
a)
C = r*ebsoleneo*A/d
= 6*(8.854*10^-12)*(1.9)/(0.0001)
=1*10^-6 F
= 1 microFarad

b)
Q = C*V
= (1*10^-6 F)*12
= 1.2*10^-5 C
= 12 micro Coulomb

c)
Surface charge density = charge/area
= 1.2*10^-5 C / 1.9 m^2
= 6.32*10^-6 C/m^2

d)
E = V/d
= 12 V / 0.0001 m
=120000 V/m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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