The plates of an air-filled parallel-plate capacitor with a plate area of 17.5 c

Question

The plates of an air-filled parallel-plate capacitor with a plate area of 17.5 cm^2 and a separation of 9.05 mm are charged to a 130-V potential difference. After the plates are disconnected from the source, a porcelain dielectric with k = 6.5 is inserted between the plates of the capacitor. What is the charge on the capacitor before and after the dielectric is inserted? Q_i = C Q_f = C What is the capacitance of the capacitor after the dielectric is inserted? F What is the potential difference between the plates of the capacitor after the dielectric is inserted? V What is the magnitude of the change in the energy stored in the capacitor after the dielectric is inserted? J

Explanation / Answer

a) Qi = C*V = (eo*A/d)*V = (8.85*10^-12*17.5*10^-4/(9.05*10^-3))*130 = 2.23*10^-10 C

Qf = Qi/k = 2.23*10^-10/(6.5) = 3.42*10^-11 C

b) C = k*Co = 6.5*(eo*A/d) = 6.5*(8.85*10^-12*17.5*10^-4/(9.05*10^-3)) = 11.12*10^-12 F

C) V = Vo/k = 130/6.5 = 20 V

D) dU = U - Uo = (k-1)*(Qi^2/(2*Co) = (6.5-1)*(2.23*10^-10)^2/(2*(8.85*10^-12*17.5*10^-4/(9.05*10^-3))) = 8*10^-8 J

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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