A dielectric-filled parallel-plate capacitor has plate area A = 20.0 cm^2, plate

Question

A dielectric-filled parallel-plate capacitor has plate area A = 20.0 cm^2, plate separation d = 5.00 mm and dielectric constant K = 5.00. The capacitor is connected to a battery that creates a constant voltage V = 10.0 V. Throughout the problem, use epsilon_0 = 8.85×10-12 C^2/N cdot m^2.

Part A: Find the energy U_1 of the dielectric-filled capacitor.

Part B: The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U_2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.

Part C: The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U_3.

Part D: In the process of removing the remaining portion of the dielectric from the disconnected capacitor, how much work W is done by the external agent acting on the dielectric?

Explanation / Answer

The capacitence of the capacitor with dielectric C _1= k*_0*A / d here k = 5.00 , d = 5*10^-3 m and A = 20*10^-4 m^2 pluging the values C_1 = 17.7 pF The capacitence of the capacitor with out dielectric C_2 = _0*A / d     = 3.54 pF The capacitence of the capacitor with out dielectric C_2 = _0*A / d     = 3.54 pF (a) potential energy U_1 = (1/2)CV^2 = (1/2) (17.7*10^-12 F)(10 V)^2 = 8.85*10^-10 j (b) the capacitence of the new capacitor with half the dielectric C = (c_1 /2 ) + (C_2 /2 ) C = (17.7 pF /2 ) + ( 3.54 pF /2 ) = 10.62 pF potential energy U_2 = (1/2)(10.62pF)( 10 V)^2 = 5.31*10^-10 j remaing question please post in another question

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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