A parallel plate, air filled capacitor, has plates of area 0.80 m^2 and a separa

Question

A parallel plate, air filled capacitor, has plates of area 0.80 m^2 and a separation of 0.040 mm. (a) Find the capacitance. (b) If a voltage of 25 volts is applied to the capacitor, what is the magnitude of the charge on each plate? (c) What is the energy stored in the capacitor? Now the plates are filled with strontium titanate having a dielectric constant of 310 and a dielectric strength of 8.0 kV/mm. (d) What is the new value of capacitance? (e) For a voltage of 25 V, what is the charge on each plate? (f) What is the energy stored in the capacitor now? (f) What is the maximum voltage that can be placed across the capacitor?

Explanation / Answer

a) area=A=0.8 m^2 separation=d=0.04 mm=4*10^(-5) m capcacitance=C=(epsilon)*A/d where epsilon=electric permitivity of air C=8.85*10^(-12)*0.8/(4*10^(-5))=1.77*10^(-7) F b)V=25 volts charge=Q=C*V=4.425*10^(-6) C c)energy=0.5*C*V^2=55.312*10^(-6) J d)new capcitance=dielectric constant*old capacitance=310*1.77*10^(-7)=54.87*10^(-6) F e) charge=25*54.87*10^(-6)=0.0014 C f) dielectric strength=8 kV/mm so maximum voltage=8 kV*0.04=320 volts

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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