Two parallel plates, each having area A = 2700cm2 are connected to the terminals

Question

Two parallel plates, each having area A = 2700cm2 are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.58cm. You may assume (contrary to the drawing) that the separation between the plates is small compared to a linear dimension of the plate.

3) A dielectric having dielectric constant = 3.6 is now inserted in between the plates of the capacitor as shown. The dielectric has area A = 2700 cm2 and thickness equal to half of the separation (= 0.29 cm) . What is the charge on the top plate of this capacitor?

What is U, the energy stored in this capacitor?

J

5)

The battery is now disconnected from the capacitor and then the dielectric is withdrawn. What is V, the voltage across the capacitor?

Explanation / Answer

C = e0 A / ( d + t(1/k - 1))

where t is the width of dielectric layer.

C = (8.854 x 10^-12 x 2700 x 10^-4 )/ (0.0029 + (0.0029/2)(1/3.6 -1))

C = 1.29 x 10^-9 F

Q= CV = 1.29 x 10^-9 x 6 = 7.74 x 10^-9 C

-----------------------------------------------------

energy stored = CV^2 /2 = (1.29 x 10^-9) x 6^2 /2 = 2.32 x 10^-8 J

-------------------------------------

charge on plates will be same.

dielectric is withdrawn ,

new capacitance C = e0 A / d = (8.854 x10^-12 x 2700 x 10^-4) / (0.0029)

C = 8.24 x 10^-10 F

V = Q / C = ( 7.74 x 10^-9 ) / ( 8.24 x 10^-10 )

V = 9.39 V

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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