A parallel plate capacitor is fabricated from two flat square metal plates 1.2 m

Question

A parallel plate capacitor is fabricated from two flat square metal plates 1.2 m on a side. A 1 mm thick mica sheet insulation if between these plates. Calculate to three significant figures the following parameters of this capacitor for a potential difference of +20 kV across the capacitor plates and assuming the dielectric constant of mica sheet insulation is K. -7. (a) Potential difference between the plates. (b) Magnitude of the electric field between the plates in MV/m units. (c) Capacitance in nF units. (d) Magnitude of charge on each plate, (e) Energy stored on the capacitor. (f) Energy density of the capacitor in kJ/mJ units calculated from the electric field E. (g) Magnitude of the real charge density on each metal plate surface. (h) Magnitude of the induced charge density on each surface of the mica insulation. (i) Assume the dielectric strength of mica insulation is 150 MV/m. At what potential difference across the capacitor plates would mica insulation likely fail by an electrical breakdown?

Explanation / Answer

g) Here, capacitance = 7 * 8.854 * 10-12 * 1.2 * 1.2/10-3

                                  = 8.924 * 10-8 F

=> real charge density = (8.924 * 10-8 * 20000)/(1.2 * 1.2)

                                     = 1.239 * 10-3 C/m2

h)   induced charge density =   1.239 * 10-3/7

                                           = 1.77 * 10-4   C/m2

i) potential difference for electrical breakdown = 150 * 106 * 10-3

                                                                        = 150000 V

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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