A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2. plate

Question

A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2. plate separation d = 10.0 mm and dielectric constant k = 4.00. The capacitor is connected to a battery that creates a constant voltage V = 7.50 V. Throughout the problem, use epsilon = 8.85 times 10^-12 C^2/N. m^2. Find the energy U_1 of the dielectric-filled capacitor. Express your answer numerically in joules. U_1 = 9.96 times 10^11 J Significant Figures Feedback: Your answer 9.95-10 11 = 9.9510~11 J was either rounded differently or used a different number of significant figures than required for this part. 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.

Explanation / Answer

a)
C = k * o A /d = 4* 8.85e-12* 0.001 / 0.001= 3.54e-11 F

U1 = Energy stored = 0.5CV^2 = 0.5*3.54e-11*7.5^2 =9.956e-10 J
U1 = 9.956e-10 J
--------------------------------------...
b)
U2 = Energy stored = [1+k] /2k* U1 = 0.625*9.956e-10
U2=6.222e-10 J
Charge residing in this new capacitor =2* U2/ V = 2*6.222e-10 /7.5
Q= 1.658e-10 couloumb.
-=====================================...
c)
After fully removing the dieelectric,
Now the capacitance C /k = 3.54e-11/4= 8.85e-12 F
U3 = Energy stored = Q^2 / (2 *8.85e-12) = 1.5381e-9 J
--------------------------------------...

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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